JEE (MAIN & ADV.

), MEDICAL + BOARD, NDA, X & IX

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INTEGRATION BY PARTS (BOARD LEVEL, XII)

SUBJECTIVE TYPE
Evaluate Q.1 to Q.33

∫ x sin x dx ∫ x sin 3x dx ∫ x sin x dx (iv) ∫ x 2e3x dx

2
1. (i) (ii) (iii)
2
(v) ∫ ( x − 1) e dx (vi) ∫ ( x + 1) e dx
2
−x x
(vii) ∫ x ⋅ 2 dx x
(viii) ∫ 2 x3e x dx
2 x3
∫ x e cos x dx 3. ∫ sec x dx
3 3
2.
log ( x + 2 )
4. (i) ∫ ( x + 2) 2
dx (ii) ∫ e x cos x dx (iii) ∫ cos −1 x dx

5. (i) ∫ log xdx (

(ii) ∫ log 1 + x 2 dx ) (iii) ∫ sin −1 xdx (iv) ∫ tan −1 x dx (v) ∫ sec−1 x dx

∫x ∫ x e dx ∫x ∫x
2 3 x 2 2
6. (i) cos x dx (ii) (iii) sin x cos x dx (iv) sin 2 x dx
x
7. (i) ∫ x sin 3xdx (ii) ∫ x cos 2 xdx (iii) ∫ x sec2 x dx (iv) ∫
1 − cos 2 x
dx

x x x − sin x
∫ x cos ∫ 1 − cos x dx
3
(v) ∫ x sin 2 cos 2 cos x dx (vi) x dx (vii)

8. (i) ∫ cos
x
2
x
dx (ii) ∫ x sin
3
xdx ( )
(iii) ∫ 1 − x 2 sin 2 x dx

∫x ∫ x cos ∫ x cot (iii) ∫ x cosec−1 x dx

2
9. cos 2 x dx 10. (i) −1
x dx (ii) −1
x dx
2 + sin 2 x x
11. (i) ∫ e x ( f ( x) + f ′ ( x ) ) dx (ii) ∫ e x sec x (1 + tan x ) dx (iii) ∫ e dx
1 + cos 2 x
 x2 + 1  2x −1  2 + sin x
(iv) ∫ e x  2 
dx (v) e 2 x  2 
dx (vi) ∫ 1 + cos x e
x/2
dx
 ( x + 1)   4x 
(1+ x ) sin −1 x
12. (i) ∫ x log ( x + 1) dx (ii) ∫ log (1 + x ) .dx (iii) ∫ x sin −1 x dx (iv) ∫ x tan −1 x dx (v) ∫ x 2 dx
log x
∫ (1 − x ) log x dx ∫ x log ( x + 3) dx ∫x (iv) ∫
2 3
13. (i) (ii) (iii) log 2 x dx dx.
xn
2 log x
∫ ( log x ) ∫x a ∫ (1 + x )
2 3x
14. (i) dx (ii) dx (iii) 2
dx

 x +1  x + sin x
15. ∫ e  ( 2 + x )2 
16. ∫ x sin 2 x cos 3x dx 17. ∫ 1 + cos x dx 18. ∫ (1 + x ) log x dx
x
dx

 
log x
19. ∫x
2
log (1 + x ) dx 20. ∫ (1 + x ) dx 3
21. ∫ (5 x + 3) 2 x − 1dx

∫x ∫x ∫x
2
22. (i) cos −1 x dx (ii) 2
cot −1 x dx (iii) 2
cosec −1 x dx

x x sin −1 x
23. (i) ∫ 1 + sin x dx (ii) ∫ sin −1 (3 x − 4 x3 )dx (iii) ∫ 1 − x2
dx

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2 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES
24. (i) Taking x as x (1/ x ) integrate it by parts. Use this to find
2 3
∫x
2
log x dx .

(ii) Taking sin x as (sin x)(1), integrate it by parts. Use this to find ∫ x cos x dx .
25. (i) ∫ e ax sin bx dx (ii) ∫ e ax cos bx dx (iii) ∫ sin(log x)dx (iv) ∫ e −x
cos 2 x cos 4 x dx

26. (i) ∫ sin 3 x dx (ii) ∫ sec −1 x dx (iii) ∫ cos x log sin x dx

sin −1 x x tan −1 x
∫x ∫ x tan x sec ∫ (iv) ∫
5 3 2
27. (i) cos x dx (ii) x dx (iii) 3
dx 3
dx
(1 − x )2 2
(1 + x )
2 2

cos x + sin x
∫ x tan
−1
28. (i) ∫ cos 2 x.log dx (ii) (2 x + 3)dx
cos x − sin x
sin −1 x − cos −1 x
∫ (sin x )
2
29. (i) −1
dx (ii) ∫ cos −1 xdx (iii) ∫ sin −1 x + cos−1 x dx
1 1  xe x
30. (i) ∫ e x ( cot x + log sin x ) dx (ii) ∫ e x  − 2  dx (iii) ∫ dx
x x  ( x + 1) 2
 1 − sin x   1 1 
(iv) ∫ e x  (v) ∫ e ( tan x + log sec x ) dx (vi) ∫ 
x
 dx −  dx
 1 − cos x   log x ( log x )2 
 
 1 
(vii) ∫ sin ( log x ) + cos ( log x ) dx (viii) ∫ log ( log x ) +  dx
2
 ( log x ) 
 x
31. ∫e
−x
( )
tan −1 e x dx . 32. ∫ log (1 + cos x ) − x tan 2  dx 33. ∫ log x.sin −1 x dx

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SOLUTION OF INTEGRATION BY PARTS (BOARD LEVEL, XII)

SUBJECTIVE TYPE
1. (i) I = ∫ x sin x dx
Integrating by parts, taking x as the first function, we get
I = x ( − cos x ) − ∫ 1 ⋅ ( − cos x ) dx = − x cos x + ∫ cos x dx = − x cos x + sin x + C

(ii) I = ∫ x sin 3 x dx
Integrating by parts, taking x as first function, we get
 − cos 3 x   − cos 3 x  x cos 3 x 1
I = x  − ∫ 1⋅   dx = − + ∫ cos 3 x dx
 3   3  3 3
x cos 3 x 1 sin 3 x x cos 3 x sin 3 x
=− + ⋅ +C = − + +C
3 3 3 3 9
(iii) I = ∫ x 2 sin x dx

Integrating by parts. Taking x 2 as Ist function, we get

I = x 2 ( − cos x ) − ∫ 2 x ( − cos x ) dx = − x 2 cos x + 2 ∫ x cos x dx

= − x 2 cos x + 2  x ( sin x ) − ∫ 1 ⋅ ( sin x ) dx 

 
= − x 2 cos x + 2 x sin x − 2 ( − cos x ) + C = − x 2 cos x + 2 x sin x + 2 cos x + C
(iv) I = ∫ x 2e3 x dx

Integrating by parts, taking x 2 as Ist function, we get

 e3 x   e3 x 
I = x2   ∫
− 2 x ⋅   dx
 3   3 
1 2 3x 2 1 2 3 x 2   e3 x   e3 x  
= x e − ∫ xe dx = x e −  x 
3  3  ∫  3  
3x
 − 1⋅   dx 
3 3 3

1 2 3x 2 3x 2 3x 1 2 2  e3 x 
= x e − xe + ∫ e dx = x 2 e3 x − xe3 x +  +C
3 9 9 3 9 9 3 
1 2 3x 2 3x 2 3x 1 2 2 
= x e − xe + e + C =  x 2 − x +  e3 x + C
3 9 27 3 9 27 
(v) I = ∫ ( x − 1) e − x dx

Integrating by parts, taking ( x − 1) as Ist function, we get

I = − ( x − 1) e− x + ∫ 1 ⋅ e − x dx = − ( x − 1) e − x − e− x + C = − xe− x + e − x − e− x + C = − xe − x + C
2
(vi) I = ∫ ( x + 1) e x dx
2
Integrating by pats, taking ( x + 1) as Ist function, we get
2
= ( x + 1) e x − ∫ 2 ( x + 1)e x dx

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2 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES
2 2
I = ( x + 1) e − 2∫ ( x + 1)e dx = e x ( x + 1) − 2 ( x + 1) e x − ∫ 1 ⋅ e x dx 
x x
 
2 2
= e x ( x + 1) − 2e x ( x + 1) + 2e x + C = e x ( x + 1) − 2 ( x + 1) + 2 + C
 
( ) (
= ex x2 + 2 x + 1 − 2 x − 2 + 2 + C = ex x2 + 1 + C )
(vii) I = ∫ x 2 x dx
Integrating by parts, taking x as the first function, we get
 2x   2x  2x 1
I = x  ∫ 
− 1 ⋅  dx = x − ∫ 2 x dx
 log 2   log 2  log 2 log 2
x2x 1  2x  x 2x 2x
= −  +C = −
log 2 log 2  log 2  log 2 ( log 2 )2

(viii) I = ∫ 2 x 3e x dx = ∫ ( 2 x ) x 2 e x dx
2 2
Put x 2 = z ⇒ 2 xdx = dz . ∴ I = ∫ z e z dz

Integrating by parts, we get, I = z.∫ e z dz − ∫  ( z ) .∫ e z dz  dz

d
 dz 
= ze z − ∫ 1.e z dz = ze z − e z + C = x 2 e x − e x + C = e x ( x 2 − 1) + C
2 2 2

3 1
2. Let, I = ∫ x 2 e x cos x 3dx Put x3 = z ⇒ 3 x 2 dx = dz ⇒ x 2 dx = dz
3
1  1 z
∴ I = ∫ e z cos z.  dz 
3∫
⇒ I= e cos z dz
3 
1 
Integrating by parts, we get, I =  e z .∫ cos z dz − ∫  ( e z ) .∫ cos z dz  dz 
d
3  dz  
1 1 1
= e z sin z − ∫ e z sin z dz  = e z sin z − ∫ e z sin z dz
3   3 3
1 1 d  
= e z sin z −  e z .∫ sin z dz − ∫  ( e z ) .∫ sin z dz dz  [Again integrating by parts]
3 3  dz  
1 1 1 1 1
= e z sin z −  e z ( − cos z ) − ∫ e z ( − cos z ) dz  = e z sin z + e z cos z − ∫ e z cos z dz
3 3  3 3 3
1 1 1 ∵ I = e z cos z dz 
⇒ I = e z sin z + e z cos z − I
3 3 3  ∫ 
 1 1 4 1 1
⇒ 1 +  I = e z [sin z + cos z ] ⇒ I = e z [sin z + cos z ] ⇒ I = e z ( sin z + cos z )
 3 3 3 3 4
1
⇒ I = e x ( sin x3 + cos x 3 ) + C
3
∵ z = x 3 
4
3. Let, I = ∫ sec3 x dx = ∫ sec x.sec 2 x dx = sec x tan x − ∫ ( sec x tan x ) tan x dx

( ) (
= sec x tan x − ∫ sec x tan 2 x dx = sec x tan x − ∫ sec2 x − 1 sec x dx )
= sec x tan x − ∫ (sec3 x dx + ∫ sec x dx = sec x tan x − I + ∫ sec x dx ∵ I = sec3 x dx 
 ∫ 
sec x tan x 1
or 2 I = sec x tan x + log | sec x + tan x | ∴ I= + log | sec x + tan x | +C
2 2

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 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES 3
log ( x + 2 ) 1 −2
4. (i) Let, I = ∫ 2
dx = ∫ log ( x + 2 ) . 2
dx = ∫ ( x + 2 ) .log ( x + 2 ) dx
( x + 2) ( x + 2)
−2 d −2 
Integrating by parts, we get, I = log ( x + 2 ) .∫ ( x + 2 ) dx − ∫  log ( x + 2 ) .∫ ( x + 2 ) dx  dx
 dx 
−2 +1 −2 +1
( x + 2) 1 ( x + 2) log ( x + 2 ) 1
= log ( x + 2 ) . −∫ . dx = − +∫ dx
( −2 + 1) x + 2 −2 + 1 ( x + 2) ( x + 2)
2

−2 +1
log ( x + 2 ) −2 log ( x + 2 ) ( x + 2)
=− + ∫ ( x + 2 ) dx = − + +C
x+2 x+2 ( −2 + 1)
log ( x + 2 ) 1
Hence, I = − − +C
x+2 ( x + 2)
(ii) Let, I = ∫ e x cos x dx …(1)

d 
Integrating by parts, we get, I = cos x.∫ e x dx − ∫  ( cos x ) .∫ e x dx  dx
 dx 
= cos xe x − ∫ ( − sin x ) e x dx = e x cos x + ∫ e x sin x dx
Again integrating by parts, we get,
d 
e x cos x + sin x.∫ e x dx − ∫  ( sin x ) .∫ e x dx  dx = e x cos x + sin xe x − ∫ cos x e x dx
 dx 
ex
⇒ I = e cos x + e sin x − I
x x
⇒ 2 I = e ( cos x + sin x )
x
⇒ I = ( sin x + cos x ) + C
2
(iii) Let, I = ∫ cos −1 x dx = ∫ 1.cos −1 x dx

d 
Integrating by parts, we get, I = cos −1 x.∫ 1.dx − ∫  ( cos −1 x ) .∫ 1.dx  dx
 dx 
−1 x
= cos −1 x. x − ∫ . x dx = x cos −1 x + ∫ dx
1 − x2 1 − x2
1
Put 1 − x 2 = z ⇒ −2 x dx = dz ⇒ x dx = − dz
2
 − 1 +1 
1  1  1 −1/2 1 z 2 
∴ I = x cos −1 x + ∫  − dz  = x cos x − ∫ z dz = x cos x −  1  + C
−1 −1

z 2  2 2  − + 1
 2 
= x cos −1 x − z1/ 2 + C = x cos −1 x − z + C
Hence, I = x cos −1 x − 1 − x 2 + c ∵1 − x 2 = z 

d 
5. (i) Let, I = ∫ log x dx = ∫ log x.1dx = (log x).∫ 1.dx − ∫  (log x) ∫ 1.dx  dx
 dx 
1 
= (log x) x − ∫  .x  dx = x log x − ∫ dx . Hence, I = x log x − x + C
x 

( ) { (
(ii) Let, I = ∫ log 1 + x 2 dx = ∫ log 1 + x 2 .1 dx = log 1 + x 2 ) } ( ) ∫ 1.dx − ∫  dxd {log (1 + x )}.∫ 1.dx  dx
2

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4 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES
2x x2
= log (1 + x 2 ) x − ∫ x dx = x log ( ) ∫ 1 + x2 dx
1 + x 2
− 2
(1 + x2 )
 1 
( )
= x log 1 + x 2 − 2 ∫ 1 − 2 
 1+ x 
(
dx = x log 1 + x 2 − 2 ∫ dx + 2 ∫ )dx
1 + x2
( )
Hence, I = x log 1 + x 2 − 2 x + 2 tan −1 x + C

(iii) Let, I = ∫ sin −1 xdx . Put sin −1 x = t ⇒ x = sin t ⇒ dx = cos tdt

∴ I = ∫ sin −1 x dx = ∫ t.cos t dt = t sin t − ∫ 1.(sin t )dt = t sin t − ∫ sin t dt = t sin t + cos t + C

Hence, I = x sin −1 x + 1 − x 2 + C ∵ cos t = 1 − sin 2 t = 1 − x 2 

 
(iv) Let, I = ∫ tan −1 x dx . Put tan −1 x = t ⇒ x = tan t ⇒ dx = sec2 t dt

∴ I = ∫ tan −1 x dx = ∫ t.sec 2 t dt

= t (tan t ) − ∫ 1. tan t dt = t (tan t ) + log | cos t | +C [Integrating by parts]

1  1 1 1 
= x tan −1 x + log +C ∵ tan t = x ⇒ cos t = = = 
1 + x2  sec t 1 + tan 2 t 1 + x2 
1
Hence, I = x tan −1 x − log |1 + x 2 | +C
2
(v) Let, I = ∫ sec−1 x dx . Put sec−1 x = t ⇒ x = sec t ⇒ dx = sec t tan t dt

∴I = ∫ t ( sec t tan t ) dt = t ( sec t ) − ∫ 1.sec t dt [Integrating by parts]

= t ( sec t ) − log sec t + tan t + C = t (sec t ) − log sec t + sec2 t − 1 + C

Hence, I = x sec −1 x − log x + x 2 − 1 + C

6. (i) Let, I = ∫ x 2 cos x dx

 d 2 
= x 2 ∫ cos x dx − ∫  x  ( ) ( ∫ cos x dx ) dx [Integrating by parts]
 dx  
= x 2 sin x − ∫ 2 x sin x dx
  d 
  dx

= x 2 sin x − 2  x ∫ sin x dx − ∫  ( x ) 

( ∫ sin x dx )dx  [Again integrating by parts]

= x 2 sin x + 2 x cos x − 2 ∫ 1.cos x dx . Hence, I = x 2 sin x + 2 x cos x − 2 sin x + C

(ii) Let, I = ∫ x3e x dx = x3e x − ∫ 3 x 2 e x dx [Integrating by parts]

{
= x 3e x − 3 x 2 e x − ∫ 2 xe x dx } [Again integrating by parts]

= x3e x − 3 x 2 e x + 6 ∫ xe x dx = x3e x − 3 x 2e x + 6  xe x − ∫ 1.e x dx  [Again integrating by parts]

 
Hence, I = x 3e x − 3 x 2e x + 6 xe x − 6e x + C
1
(iii) Let, I = ∫ x 2 sin x cos x dx = ∫ x 2 .2 sin x cos x dx
2

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 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES 5

1 1   − cos 2 x    − cos 2 x   
= ∫ x 2 sin 2 xdx =  x 2   − ∫ 2 x    dx  [Integrating by parts]
2 2  2    2  
1  − x 2 cos 2 x 
=  + ∫ x cos 2 x dx 
2 2 
1  − x 2 cos 2 x  sin 2 x sin 2 x  
=  +  x. −∫ dx   [Again integrating by parts]
2 2  2 2 
1  − x 2 cos 2 x x sin 2 x cos 2 x  − x 2 cos 2 x x sin 2 x cos 2 x
=  + + +C . Hence, I= + + +C
2 2 2 4  4 4 8
(1 − cos 2 x) 1 1
(iv) Let, I = ∫ x 2 sin 2 xdx = ∫ x 2 dx =  ∫ x 2 (1 − cos 2 x)dx  =  ∫ x 2 dx − ∫ x 2 cos 2 x dx 
2 2   2 
1  x3  2 sin 2 x sin 2 x 
=  − x . − ∫ 2 x. dx  [Integrating by parts]
2 3  2 2 
1  x3 x 2 sin 2 x 
=  − + ∫ x sin 2 x dx 
2 3 2 
1  x 3 x 2 sin 2 x   − cos 2 x   − cos 2 x  
=  − + x  − ∫  dx  [Again integrating by parts]
2 3 2   2   2  
x3 x 2 sin 2 x x cos 2 x sin 2 x
Hence, I = − − + +C
6 4 4 8
 d ( x) 
7. (i) Let, I = ∫ x sin 3 x dx = x ∫ sin 3 x dx − ∫ 
 dx 
 ( ∫ sin 3x dx ) dx [Integrating by parts]

 cos 3 x    cos 3 x   − x cos 3x 1 x 1  sin 3x 

= x.  −  − ∫ 1 −   dx = + ∫ cos 3x dx = − cos 3x +  +C
 3    3  3 3 3 3 3 
x 1
Hence, I = − cos 3 x + sin 3 x + C
3 9
 sin 2 x   sin 2 x 
(ii) Let, I = ∫ x cos 2 x dx = x.   − ∫ 1.  dx [Integrating by parts]
 2   2 
x sin 2 x cos 2 x
Hence, I = + +C
2 4
(iii) Let, I = ∫ x sec2 x dx = x. ( tan x ) − ∫ (1. tan x ) dx [Integrating by parts]

= x tan x − ∫ tan x dx = x tan x + log | cos x | +C

Hence, I = x tan x + log | cos x | +C
x x 1
(iv)Let, I = ∫ dx = ∫ 2
dx = ∫ x cos ec 2 x dx
1 − cos 2 x 2 sin x 2
1
= x ( − cot x ) − ∫ {1.(− cot x)}dx  [Integrating by parts]
2 
1  1 
Hence, I = − [ x cot x + ∫ cot x dx] = − x cot x + log | sin x |  + C
2  2 
x x 1  x x 1
(v) Let, I = ∫ x sin cos cos x dx = ∫ x.  2sin cos  cos x dx = ∫ x sin x cos x dx
2 2 2  2 2 2

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6 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES
1 1
=
4 ∫ x.(2sin x cos x)dx = ∫ x sin 2 x dx
4
1   − cos 2 x    cos 2 x   
=  x.   + ∫ 1.    dx  [Integrating by parts]
4  2    2  
1  − x cos 2 x 1  1  − x cos 2 x 1  − x cos 2 x sin 2 x
=  + ∫ cos 2 x dx  Hence, I =  + sin 2 x  + C = + +C
4 2 2  4 2 4  8 16

(vi) Let, I = ∫ x cos3 x dx = ∫ x

( cos 3x + 3cos x ) dx [∵ cos 3 x = 4 cos3 x − 3cos x]
4
1 1  ( sin 3x )  sin 3x  
=
4  ∫ x cos 3 x dx + 3∫ x cos x dx  =  x.
 4 3
− ∫ 1.
 3 
{
 dx + 3 x sin x − ∫ (1.sin x )dx 

}
1  x sin 3 x 1 
=  − ∫ sin 3x dx + 3 x sin x − 3∫ sin x dx 
4 3 3 
1  x sin 3x 1 
Hence, I =  + cos 3x + 3x sin x + 3cos x  + C
4 3 9 
x − sin x x sin x
(vii) Let, I = ∫ dx = ∫ dx − ∫ dx
1 − cos x 1 − cos x 1 − cos x
x x
2sin cos  x
x 2 x 2 2 dx 2
= ∫ cosec dx − ∫
2 2 2 x
∵1 − cos x = 2sin 2 
2sin
2
1 x x
= ∫ x cosec 2 dx − ∫ cot dx
2 2 2
1  x   x   x
=  x  −2 cot  − ∫ 1.  −2 cot   dx  − ∫ cot dx [Integrating by parts]
2  2   2   2
x x x x
= − x cot + ∫ cot dx − ∫ cot dx . Hence , I = − x cot + C
2 2 2 2
x
8. (i) Let, I = ∫ dx ∴ I = ∫ x sec 2 x dx.
cos 2 x
d 
⇒ I = x ∫ sec 2 xdx − ∫  ( x).∫ sec 2 x dx  dx [Integrating by parts]
 dx 
= x tan x − ∫ 1.tan x dx Hence, I = x tan x + log | cos x | + C

(ii) Let, I = ∫ x sin 3 x dx

3 1
We have, sin 3 x = 3sin x − 4sin 3 x ∴ sin 3 x = sin x − sin 3 x
4 4
3 1  3 1
∴ I = ∫ x  sin x − sin 3 x  dx = ∫ x sin x dx − ∫ x sin 3 x dx
4 4  4 4
3 1   cos 3 x   cos 3 x  
= x ( − cos x ) − ∫ 1(− cos x)dx  −  x  −  − ∫ 1 −  dx
4  4  3   3  
3 1 x cos 3 x sin 3 x 
Hence, I = [ − x cos x + sin x ] −  − + +C
4 4 3 9 

( )
(iii) Let, I = ∫ 1 − x 2 sin 2 x dx

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 − cos 2 x   − cos 2 x 
⇒ I = (1 − x 2 )   − ∫ ( −2 x )   dx [Integration by parts]
 2   2 
x2 − 1 x2 − 1  sin 2 x sin 2 x 
= cos 2 x − ∫ x cos 2 x dx = cos 2 x −  x − ∫ 1. dx 
2 2  2 2 
x2 −1 x 1  cos 2 x  x 2 − 1 x 1
= cos 2 x − sin 2 x +  − = cos 2 x − sin 2 x − cos 2 x
2 2 2 2  2 2 4
2x2 − 3 x
Hence, I = cos 2 x − sin 2 x + C
4 2
 1 + cos 2 x  1 2 1 2
9. Let, I = ∫ x 2 cos 2 x dx = ∫ x 2  dx = ∫ x dx + ∫ x cos 2 x dx
 2  2 2
1 x 3 1   sin 2 x   sin 2 x  
= . +  x2   − ∫ 2x   dx  [Integrating by parts]
2 3 2  2   2  
x 3 x 2 sin 2 x 1
= + − ∫ x sin 2 x dx
6 4 2
x3 x 2 sin 2 x 1   cos 2 x   − cos 2 x  
= + − x −  − ∫ 1.   dx  [Again integrating by parts]
6 4 2  2   2  
x 3 x 2 sin 2 x x cos 2 x 1 x 3 x 2 sin 2 x x cos 2 x 1
= + + − ∫ cos 2 x dx . Hence, I = + + − sin 2 x + C
6 4 4 4 6 4 4 8
10. (i) Let, I = ∫ x cos −1 x dx ∴ I = ∫ cos −1 x.xdx

x2 −1 x 2
⇒ I = cos −1 x. −∫ . dx [Integrating by parts]
2 1 − x2 2
x2 1 x2 x2
⇒ I= cos −1 x + ∫ dx Let, I1 = ∫ dx . Put x = cos θ ∴ dx = − sin θ dθ
2 2 1 − x2 1 − x2
cos 2 θ 1 + cos 2θ 1 1 sin 2θ
∴ I1 = ∫ .(− sin θ )dθ = − ∫ cos 2 θ dθ = − ∫ dθ = − θ − .
2
1 − cos θ 2 2 2 2

1 sin θ cos θ 1 cos θ 1 − cos 2 θ 1 x 1 − x2

=− θ− =− θ− = − cos −1 x −
2 2 2 2 2 2
Putting the value of I1 in I , we get,

x2 1 1 x 1 − x2  x2 1 1
−1 −1
I = cos x +  − cos x − +C = cos −1 x − cos −1 x − x 1 − x 2 + C
2 2  2 2 
2 4 4

Alternatively : Let, I = ∫ x cos −1 x dx . Put x = cos θ ∴ dx = − sin θ dθ

1
∴ I = ∫ cos θ .θ (− sin θ dθ ) = −
2∫
θ sin 2θ dθ

1  − cos 2θ − cos 2θ 
⇒ I = − θ . − ∫ 1. dθ  [Integrating by parts]
2 2 2 
θ cos 2θ 1 θ cos 2θ 1 sin 2θ
= − ∫ cos 2θ dθ = − . +C
4 4 4 4 2
θ ( 2 cos 2 θ − 1) 1 θ cos 2 θ θ cos θ 1 − cos 2 θ
= − .2 sin θ cos θ + C = − − +C
4 8 2 4 4

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Hence, I =
( cos x ) x
−1 2

−
cos −1 x x 1 − x 2
− +C
2 4 4
(ii) Let, I = ∫ x cot −1 xdx ⇒ I = ∫ cot −1 x.xdx

x2 −1 x 2
Integrating by parts, we get, I = ∫ x cot −1 x dx = cot −1 x − ∫ . dx
2 1 + x2 2
x 2 cot −1 x 1 x2 x 2 cot −1 x 1  1  x 2 cot −1 x 1
=
2
+ ∫
2 1+ x 2
dx =
2
+ ∫  1 −
2  1+ x 2 
dx =
2
+  x − tan −1 x  + C
2

=
x 2 cot −1 x 1 1π −1 
+ x −  − cot x  + C =
x 2 + 1 cot −1 x 1 π
+ x− +C
( )
2 2 2 2  2 2 4

=
(x 2
)
+ 1 cot −1 x
+
1
x + C1 , where C1 = C −
π
2 2 4
(iii) Let, I = ∫ x cosec −1 x dx ⇒ I = ∫ cosec −1 x.xdx

x2 −1 x2
Integrating by parts, we get, I = cosec x. − ∫ . dx −1

2 x x2 −1 2
x 2 cosec −1 x 1 x
⇒ I= + ∫ dx …(i)
2 2 2
x −1
x 1 dz 1 z1/ 2
Let, I1 = ∫ dx . Put z = x 2 − 1 ∴ dz = 2 x dx ∴ I1 = ∫ . = . = z = x2 − 1
x2 − 1 z 2 2 1/ 2
x 2 cosec−1 x 1 2
∴ Putting the value of I1 in (i), we get, I = + x −1 + C
2 2
11. (i) Let, I = ∫ e x ( f ( x ) + f ′ ( x ) ) dx = ∫ f ( x ) .e x dx + ∫ e x f ′ ( x ) dx

=  f ( x ) e x − ∫ f ′ ( x ) e x dx  + ∫ e x f ′ ( x ) dx = f ( x ) e x + C
 
(ii) Let, I = ∫ e x sec x (1 + tan x) ) dx

∴ I = ∫ e x ( sec x + sec x tan x ) dx ⇒ I = ∫ sec x.e x dx + ∫ e x sec x tan dx

Evaluating first integral of I by parts, we get,
I = sec x.e x − ∫ sec x tan x.e x dx  + ∫ e x sec x tan xdx = sec x.e x + C
 
2 + sin 2 x x
(iii) Let, I = ∫ e dx
1 + cos 2 x
2 + 2 sin x cos x x  1 sin x  x
∴ I =∫ 2
e dx = ∫  2
+  e dx = ∫ (sec x + tan x)e dx
2 x

2 cos x  cos x cos x 

⇒ I = ∫ tan x e x dx + ∫ sec 2 xe x dx

Evaluating first integral of I by parts, we get, I =  tan xe x − ∫ sec 2 x.e x dx  + ∫ sec2 x.e x dx = tan x.e x + C
 
 x2 + 1 
(iv)Let, I = ∫ e x   dx
 ( x + 1)2 
 
 x2 − 1 + 2   x2 −1 2  
x x −1 2 
∴ I = ∫ ex   dx = ∫ e x  +  dx = ∫  x + 1 ( x + 1)2  dx
e  +
 ( x + 1)2  2 2
 ( x + 1) ( x + 1) 
   
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 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES 9
d  x − 1  ( x + 1) .1 − ( x − 1) .1 2
We know that   = 2
= 2
dx  x + 1  ( x + 1) ( x + 1)
x −1 x 2
∴ We can write I = ∫ e dx + ∫ e x . 2
dx
x +1 ( x + 1)
 x −1 2  2 x −1 x
Evaluating first integral of I by parts, we get I =  ex − ∫ 2
.e x
dx  + ∫ ex . 2
dx = e +C
 x + 1 ( x + 1)  ( x + 1) x +1

 2x −1 
(v) Let, I = ∫ e2 x  2 
dx. Put z = 2 x ∴ dz = 2dx
 4x 
z − 1  dz 1 z  1 1  1 1 z 1 1
∴ I = ∫ e z  2 
= ∫ e  − 2  dz ⇒ I= ∫ e dz − ∫ e z . 2 dz
 z  2 2 z z  2 z 2 z
Evaluating first integral of I by parts, we get,
1 1 z 1 z  1 z 1 11 z e2 x
I=  e − ∫ − 2 e dz  − ∫ e . 2 dz =  e  + C = +C
2z z  2 z 2 z  4x
2 + sin x x /2
(vi) Let, I = ∫ e dx
1 + cos x
x x  x
2 + 2sin cos sin 
 2 e x / 2 dx =  sec 2 + tan  e x /2 dx.
1 x x
∴ I=
∫ 2
x
2 e x / 2 dx =
∫ 2x
 +
x 
 2 ∫ 2
2 cos 2  cos cos 
2  2 2
1
Let, z =
x
2
(
∴ dz = dx . ∴ I = 2 ∫ sec 2 z + tan z e z dz
2
)
∴ We can write I = 2 ∫ tan z e z dz + 2 ∫ sec 2 ze z dz
Evaluating first integral of I by parts, we get,
x
I = 2  tan z.e z . − ∫ sec 2 z.e z dz  + 2 ∫ sec 2 ze z dz = 2 tan z.e z + C = 2 tan .e x / 2 + C
  2
12. (i) Let, I = ∫ x log ( x + 1) dx = ∫ log ( x + 1) .x dx
x2  1 x2 
[Integrating by parts]
= log ( x + 1) . −∫ .  dx
2  x +1 2 
x2 1 x2 x2 1 x2 −1 + 1
= log( x + 1) − ∫ dx = log( x + 1) − ∫ dx
2 2 x +1 2 2 x +1
x2 1  x2 −1 1  x2 1  1  
= log( x + 1) − ∫  +  dx = log ( x + 1) −  ∫  ( x − 1) +  dx
2 2  x +1 x +1  2 2  x + 1  

x2 1  x2 
= log ( x + 1) −  − x + log x + 1  + C
2 2 2 
x2 x2 x 1
Hence, I = log ( x + 1) − + − log x + 1 + C =
( x 2 − 1) x2 x
log ( x + 1) − + + C
2 4 2 2 2 4 2
(1+ x )
(ii) Let, I = ∫ log (1 + x ) dx = ∫ (1 + x ) log (1 + x ) dx ∵ log m n = n log m 

 1   x 2  
= log (1 + x ) ∫ (1 + x ) dx − ∫    x +   dx [Integrating by parts]
 1 + x   2  

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10 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES
 x   1 x ( x + 2) 
2
 x2  1 x2 + 2x
= log (1 + x )  x +  − ∫  .  dx = log (1 + x )  + x − ∫ dx
 2  1 + x 2   2  2 x +1
 x2 + 2 x  1 x2 + 2 x + 1 −1  x2 + 2x  1  1 
= log (1 + x )   − ∫ dx = log (1 + x ) .   − ∫  x +1−  dx
 2  2 x +1  2  2  x +1

=
(x 2
+ 2x ) log (1 + x ) − x x log (1 + x )
− +
2
+ C Hence, I =
2
( x + 1) log(1 + x) − x 2 − x + C
2 4 2 2 2 4 2
x2  1 x 2 
(iii) Let, I = ∫ x sin −1 x dx = (sin −1 x) −∫ .  dx
2  1 − x 2 2 
x2 1 − x2 x2 1 1 − x2 − 1
= .sin −1 x + ∫ dx = .sin −1 x + ∫ dx
2 2 1 − x2 2 2 1 − x2
x2 1  1 − x 2 1  x 2 1  1 
= .sin −1 x +  ∫ dx − ∫ dx  = .sin −1 x +  ∫ 1 − x 2 dx − ∫ dx 
2 2  1 − x 2 1 − x 2  2 2  1 − x 2 
x2 1  1 1  
= .sin −1 x +   x 1 − x 2 + sin −1 x  − sin −1 x  + C
2 2  2 2  
1 2 −1 1 1
Hence, I = x sin x + x 1 − x 2 − sin −1 x + C
2 4 4
Alternatively : Let, I = ∫ x sin −1 x dx . Put sin −1 x = t ⇒ x = sin t ⇒ dx = cos t dt
1
∴ I = ∫ sin t.t.cos t dt = t ( 2sin t cos t ) dt
2∫
1 1   − cos 2t   − cos 2t   1  −t cos 2t 1  sin 2t  
=
2 ∫ t sin 2t dt = t 
2  2 
 − ∫ 1. 
 2   2
 dt  = 
2
+ 
2  2  
 +C

t cos 2t sin 2t t (1 − 2 sin 2 t ) 2sin t 1 − sin 2 t sin −1 x (1 − 2 x 2 ) 2 x 1 − x 2

=− + +C = − + +C = − + +C
4 8 4 8 4 4
d 
(iv) Let, I = ∫ x tan −1 x dx = ( tan −1 x ).∫ x dx − ∫  ( tan −1 x ) .∫ xdx  dx
 dx 
x2  1 x2  x 2 tan −1 x 1  1 
(
= tan −1 x . ) 2
−∫ 2
.  dx =
2
− ∫ 1 −  dx
2  1 + x2 
 (1 + x ) 2 
x 2 tan −1 x 1 1 1 x 2 tan −1 x x 1
= − ∫ dx + ∫ 2
dx = − + tan −1 x + C
2 2 2 1+ x 2 2 2
1 1
( )
Hence, I = 1 + x 2 tan −1 x − x + C
2 2
sin −1 x
(v) Let, I = ∫ dx . Put sin −1 x = z ⇒ x = sin z ⇒ dx = cos z dz
x2
z cos z dz
∴I =∫ = ∫ z ( cosec z cot z ) dz
sin 2 z
 d ( z) 
= z ∫ cosec z cot z dz − ∫  .∫ cosec z cot z )  dz [Integrating by parts]
 dz 
= z ( − cosec z ) − ∫ 1. ( − cosec z ) dz = − z cosec z + ∫ cosec z dz

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− sin −1 x 1 1 − x2
= − z cosec z + log | cosec z − cot z | +C = + log − +C
x x x

sin −1 x 1 − 1 − x2
Hence, I = − + log +C
x x

13. (i) Let, I = ∫ (1 − x 2 ) log x dx

d 
Integrating by parts, we get, I = log x.∫ (1 − x 2 )dx − ∫  ( log x ) .∫ (1 − x 2 ) dx  dx
 dx 
 x3  1 x3   3x − x3   x2 
= log x  x −  − ∫  x −  dx =   log x − ∫ 1 −  dx
 3 x 3  3   3
 3x − x3  1 2  3x − x3  1 x3  3x − x3  x3
=
 3 
 log x − ∫ 1.dx +
3∫
x dx = 
 3 
 log x − x + .
3 3
+ C =
 3 
 log x − x + + C
9

(ii) Let, I = ∫ x log ( x + 3) dx

d 
Integrating by parts, we get, I = log ( x + 3) ∫ xdx − ∫  ( log ( x + 3) ) ∫ xdx  dx
 dx 
x2 1  x2  x2 1 x2
= log ( x + 3) . −∫ ( )
2 ∫ x+3
dx = log x + 3 − dx.
2 x + 3  2  2
x2 1  3x  x2 1 3 x
= log ( x + 3) − ∫  x −  dx = log ( x + 3) − ∫ x dx + ∫ dx
2 2  x + 3 2 2 2 x+3
x2 1 x 2 3 ( x + 3 − 3) x2 1 3 3 
= log ( x + 3) − . + ∫ dx = log ( x + 3) − x 2 +  ∫ 1.dx − ∫ dx 
2 2 2 2 x+3 2 4 2 x+3 
x2 1 3 9 1 x2 1 3 9
= log ( x + 3) − x 2 + x − ∫ dx = log ( x + 3) − x 2 + x − log ( x + 3) + C
2 4 2 2 x+3 2 4 2 2

=
(x 2
− 9)
log ( x + 3) −
x 2 3x
+ +C
2 4 2
(iii) Let, I = ∫ x3 log 2 xdx.

d 
Integrating by parts, we get, I = log 2 x.∫ x 3dx − ∫  ( log 2 x ) .∫ x 3dx  dx
 dx 
x4 1 x4 1 1 1 1 x4 1 1
= log 2 x. − ∫ ( 2 ) . dx = x 4 log 2 x − ∫ x 3 dx = x 4 log 2 x − . + C = x 4 log 2 x − x 4 + C
4 2x 4 4 4 4 4 4 4 16
log x
(iv) Let, I = ∫ n dx = ∫ x − n log x dx
x
d 
Integrating by parts, we get, I = log x ∫ x − n dx − ∫  ( log x ) .∫ x − n dx  dx
 dx 
x − n +1 1 x − n +1 1 1
= log x.
( − n + 1 )
−∫ .
x ( − n + 1)
dx =
( )
1 − n
log x. x − n +1 −
( )
1 − n ∫ x − n dx

log x 1 1 x − n +1 log x 1 1 1  1 
= . − +C = − . n −1 + C =  log x − +C
(1 − n ) x n −1
(1 − n ) ( −n + 1) (1 − n ) x n −1 2
(1 − n ) x (1 − n ) x n −1  (1 − n ) 

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12 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES
2 2
14. (i) Let, I = ∫ ( log x ) dx = ∫ 1. ( log x ) dx
Integrating by parts, taking unity as the second function, we get,
2 d 2  2 1
I = ( log x ) .∫ 1.dx − ∫  ( log x ) .∫ 1.dx dx = ( log x ) . x − ∫ 2 log x. . x dx
 dx  x
2 2
= x ( log x ) − 2 ∫ log x dx = x ( log x ) − 2∫ 1.log x dx

2  d  
= x ( log x ) − 2 log x.∫ 1.dx − ∫  ( log x ) .∫ 1dx  dx  [Again integrating by parts]
  dx  
 1  2 2
= x ( log x ) − 2 log x. x − ∫ . x dx  = x ( log x ) − 2 x log x + 2 ∫ 1.dx = x ( log x ) − 2 x log x + 2 x + C
2

 x 
(ii) Let, I = ∫ x 2 a 3 x dx

d 
Integrating by parts, we get, I = x 2 ∫ a 3 x dx − ∫  ( x 2 ) .∫ a 3 x dx  dx
 dx 
a 3x a3x x2a3x 2
− ∫ 2 x.
3log a 3log a ∫
= x2. dx = − x.a3 x dx
3log a 3log a
x 2a 3 x 2  d  
= −  x.∫ a 3 x dx − ∫  ( x ) .∫ a 3 x dx  dx  [Again integrating by parts]
3log a 3log a   dx  
x2a3x 2  a3x a3 x  x 2 a 3 x 2 x a3 x 2
= −  x. − ∫ 1. dx  = − 2
+ 2 ∫
a 3 x dx
3log a 3log a  3log a 3log a  3log a 9 ( log a ) 9 ( log a )

x2a3x 2 3x 2 a3 x x2a3x 2 xa 3 x 2a 3 x
= − xa + . + C = − + +C
3log a 9 ( log a )2 2
9 ( log a ) 3log a 3log a 9 ( log a )2 27 ( log a )3
log x 1
(iii) Let, I = ∫ 2
dx = ∫ log x. dx
(1 + x ) (1 + x )
2

1  d 1 
Integrating by parts, we get, I = log x.∫ 2
dx − ∫  ( log x ) .∫ 2
dx  dx
(1 + x )  dx (1 + x ) 
−2 +1 −2 +1

= log x
(1 + x ) 1 (1 + x )
−∫ .
−1 1 −1
dx = − log x. (1 + x ) + ∫ (1 + x ) dx
−2 + 1 x ( −2 + 1) x

1 1 log x 1 1   1 1 1+ x − x 1 
= − log x. +∫ dx = − +∫ − dx ∵ − = = 
(1 + x ) x (1 + x ) (1 + x )  x 1 + x   x 1 + x x (1 + x ) x (1 + x ) 
log x 1 1 log x
=− + ∫ dx − ∫ dx = − + log x − log 1 + x + C
(1 + x ) x 1+ x (1 + x )
log x x  m
=− + log +C ∵log m − log n = log n 
(1 + x ) 1+ x
 x +1   x + 2 −1
15. Let, I = ∫ e x  2 
dx = ∫ e x   dx
 (2 + x)   ( 2 + x )2 
   
 x+2 1  1 1
= ∫ ex  −  dx = ∫ e .
x
dx − ∫ e x . dx
 ( x + 2 ) 2 ( x + 2 )2 
  ( x + 2) ( x + 2)
2

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 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES 13
Integrating the first integral by parts, we get,
 1  x d  1  x  1
I =  .∫ e dx − ∫    .∫ e dx  dx − ∫ e .
x
2
dx
 x+2  dx  x + 2   ( x + 2)
 1  x  1  x 1
=  e − ∫  − 2 

e dx − ∫ e x 2
dx
 x+2  ( x + 2 )  ( x + 2 )
 1  1 1 ex
= ex   ∫
+ − ∫ ( x + 2 )2 = +C
x x
e 2
dx e dx
 x+2 ( x + 2) x+2
1
Let, I = ∫ x sin 2 x cos 3 x dx =
2∫
16. x.cos 3 x sin 2 x dx

1 1 1
= .∫ x ( sin 5 x − sin x ) dx = ∫ x sin 5 xdx − x sin x dx …(1)
2 2 2
 cos 5 x   cos 5 x  x cos 5 x 1
Now, ∫ x sin 5 x dx = x  −5 
 − ∫ 1.  −
 5 
 dx = −
5
+ ∫ cos 5 x dx
5
x cos 5 x 1 sin 5 x x cos 5 x sin 5 x
=− + . =− + …(2)
5 5 5 5 25
Again, ∫ x sin x dx = x ( − cos x ) − ∫ 1. ( − cos x ) dx
= − x cos x + ∫ cos x dx = − x cos x + sin x …(3)
1  x cos 5 x sin 5 x  1
From (1), (2) & (3), I =  − +  − ( − x cos x + sin x ) + C
2 5 25  2
x cos 5 x sin 5 x x cos x sin x
Hence, I = − + + − +C
10 50 2 2
x x
2 sin cos
x + sin x x sin x x 2 2 dx
17. I =∫ dx = ∫ dx + ∫ dx = ∫ dx + ∫
1 + cos x 1 + cos x 1 + cos x 2 cos 2 x
2 cos 2 x

2 2
 x x 
 tan tan
1 x x 1 2 − 1. 2 dx  + tan x dx
= ∫ x.sec 2 dx + ∫ tan dx =  x.
2 2 2 2 1 ∫ 1  ∫ 2

 2 2 
x x x x
= x tan − ∫ tan dx + ∫ tan dx = x tan + C
2 2 2 2
18. Let, I = ∫ (1 + x ) log x dx
2 2

Integrating by parts, taking log x as first function, we get, I=

(1 + x ) log x − ∫
(1 + x ) 1
. dx
2 2 x
2
1 2 1 1 + 2x + x 1 2 1 1 
= (1 + x ) log x − ∫ dx = (1 + x ) log x − ∫  + 2 + x dx
2 2 x 2 2 x 
1 2 1 x2   x2   x2 
= (1 + x ) log x −  log x + 2 x +  + C =  x +  log x −  x +  + C
2 2 2  2   4
x3 1 x3
19. Let, I = ∫ x 2 log (1 + x ) dx = log (1 + x ) . −∫ . dx
3 1+ x 3

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14 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES
3 3
x 1 x
⇒ I = log (1 + x ) − ∫ dx …(1)
3 3 1+ x
x3 x3 + 1 − 1 x3 + 1 dx
Now, let I1 = ∫ 1+ x ∫ x +1
dx = dx = ∫ x + 1 dx − ∫ x + 1
x3 x2
=∫ ( )
x 2 − x + 1 dx − ∫
dx
= − + x − log ( x + 1)
x +1 3 2
x3 1  x3 x 2 
On putting the value of I1 in (1), we get, I = log (1 + x ) −  − + x − log (1 + x )  + C
3 3 3 2 
 x3 + 1  x3 x2 x
⇒ I =  log (1 + x ) − + − +C
 3  9 6 3
log x
20. Let I = ∫ 3
dx . On integrating by parts taking as u , we get,
(1 + x )
 1   1  1 log x 1 dx
I = ( log x ) .  −  − ∫  −  dx = − . + ∫ …(1)
 2 (1 + x )   2 (1 + x ) x 
2 2
2 (1 + x ) 2 (1 + x )2 x
2

1 1+ x − x 1 1 1+ x − x 1 1 1 1
Now, = = − = − = − −
x (1 + x )
2
x (1 + x )
2
x (1 + x ) (1 + x ) 2
x (1 + x ) (1 + x ) 2
x 1 + x (1 + x )2

dx 1 1 1  1 x 1
∴ ∫ x (1 + x )2 ∫  x 1 + x (1 + x )2  dx = log x − log (1 + x ) + 1 + x = log 1 + x + 1 + x
=  − −
 
1 log 1 x 1
∴ From (1), I = − + log + +C
2
2 (1 + x ) 2 1 + x 2 (1 + x )

3  3 

21. Let, I = ∫ ( 5 x + 3) 2 x − 1dx = ( 5 x + 3) .

( 2 x − 1) 2  ( 2 x − 1) 2 
− ∫ 5.
3 
dx
3
2.  2. 
2  2 
5
1 5 ( 2 x − 1) 2
3
1 3
1 5
= ( 5 x + 3)( 2 x − 1) − . 2 + C = ( 5 x + 3)( 2 x − 1) 2 − ( 2 x − 1) 2 + C
3 3 5 3 3
2.
2
1 3
1 3
= ( 2 x − 1) 2 5 x + 3 − ( 2 x − 1) + C = ( 2 x − 1) 2 ( 3x + 4 ) + C
3 3
22. (i) Let I = ∫ x 2 cos −1 x dx ∴ I = ∫ cos −1 x.x 2 dx
x3 −1 x 3 x 3 cos −1 x 1 x3
Integrating by parts, we get, I = cos −1 x. −∫ . dx = + ∫ dx
3 1 − x2 3 3 3 1 − x2

x3 x2
Let I1 = ∫ dx ∴ I1 = ∫ .x dx . Put z = 1 − x 2 ∴ dz = −2 xdx
2 2
1− x 1− x
1 − z dz 1
∴ I1 = ∫ . = − ∫ z −1/2 − z1/2 )dz
z −2 2
1 2 z 3/2   z   1 − x2   x2 + 2 
= −  2 z1/2 − =
  − 1  z =  − 1  1 − x 2
= −   1− x
2

2 3  3   3   3 
Putting the value of I1 in I , we get,

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 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES 15

x3 cos −1 x 1   x 2 + 2   x 3 cos −1 x 1 2
I= + −  2
 1− x  + C = − ( x + 2) 1 − x 2 + C
3 3  3   3 9

(ii) I = ∫ x 2 cot −1 x dx ∴ I = ∫ cot −1 x.x 2 dx

x3 −1 x 3 x 3 cot −1 x 1 x3
−∫
3 ∫ 1 + x2
Integrating by parts, we get, I = cot −1 x. . dx = + dx
3 1 + x2 3 3
x 3 cot −1 x 1  x  x 3 cot −1 x 1  x 2 1 
= + ∫ x − 2 
dx = +  − log |1 + x 2 | + C
3 3  1+ x  3 3 2 2 
x 3 cot −1 x 1 2 1
=
3 6 6
(
+ x − log 1 + x 2 + C )
(iii) I = ∫ x 2 cosec−1 x dx ⇒ I = ∫ cosec −1 x.x 2 dx

x3 cosec −1 x 1 x2
Integrating by parts, we get, I = + ∫ dx
3 3 x2 −1
x2
Let I1 = ∫ dx Put x = cosec θ ⇒ dx = − cosec θ cot θ dθ
x2 − 1
cosec 2 θ
∴ I1 = − ∫ .cosec θ cot θ dθ = − ∫ cosec3 θ dθ = − ∫ cosec θ .cosec 2 θ dθ
2
cosec θ − 1
= −  − cosec θ cot θ − ∫ ( − cosec θ cot θ )( − cot θ ) dθ  = cosec θ cot θ + ∫ cosec θ cot 2 θ dθ
 
( )
= cosec θ cot θ + ∫ cosec 2 θ − 1 cosec θ dθ

⇒ I1 = cosec θ cot θ − I1 − ∫ cosec θ dθ ⇒ 2 I1 = cosec θ cot θ − log | cosec θ − cot θ |

1 1 1 1
∴ I1 = cosec θ cot θ − log | cosec θ − cot θ | ⇒ I1 = x x 2 − 1 − log x − x 2 − 1
2 2 2 2
x3 cosec −1 x 1  1 1 
Putting the value of I1 in I , we get, I = +  x x 2 − 1 − log x − x 2 − 1  + C
3 3 2 2
x 3 cosec−1 x x x 2 − 1 1
= + − log x − x 2 − 1 + C
3 6 6
x
23. (i) Let, I = ∫ dx
1 + sin x
x x dx 1 π x 
∴ I =∫ dx = ∫ = ∫ x.sec2  −  dx
π  π x  2  4 2
1 + cos  − x  2 cos 2  − 
2   4 2
 π x  π x  
 x tan  −  1.tan  −  
Integrating by parts, we get, I = 1   4 2 −  4 2  dx 
2 1 ∫ 1 
− −
 2 2 

π x
log sec  − 
π x   4 2 π x  π x 
= − x tan  −  + + C = − x tan  −  − 2 log sec  −  + C
 4 2 −
1  4 2  4 2
2
(ii) Let, I = ∫ sin −1 ( 3 x − 4 x 3 ) dx. Put x = sin θ ∴ dx = cos θ dθ

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16 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES
∴ I = ∫ sin ( 3sin θ − 4sin θ ) .cos θ dθ = ∫ sin −1 ( sin 3θ ) cos θ dθ = ∫ 3θ cos θ dθ = 3∫ θ .cos θ dθ
−1 3

Integrating by parts, we get,

I = 3 θ sin θ − ∫ sin θ dθ  = 3θ sin θ + 3cos θ + C = 3x sin −1 x + 3 1 − x 2 + C
 
x sin −1 x sin θ .θ .cos θ dθ
(iii) Let, I = ∫ dx. Put x = sin θ ∴ dx = cos θ dθ ∴I =∫ = ∫ θ .sin θ dθ
2
1− x 1 − sin 2 θ
Integrating by parts, we get,
I = θ ( − cos θ ) − ∫ 1( − cos θ ) dθ = − θ cos θ + sin θ + C = − sin −1 x. 1 − x 2 + x + C

1 1
(i) We have x 2 = x 3   ∫ x dx = ∫ x
2 3
24. ∴   dx
x x
x3 x3
⇒ = x 3 .log | x | − ∫ 3 x 2 log | x | dx ⇒ = x 3 log x − 3∫ x 2 log x dx
3 3
x3 x3 log x x 3
⇒ 3∫ x 2 log x dx = x3 log x − ⇒ ∫ x 2 log x dx = − +C
3 3 9
(ii) We have sin x = ( sin x )(1) ∴ ∫ sin x dx = ∫ ( sin x ) . (1) dx
⇒ − cos x = ( sin x ) x − ∫ ( cos x ) x dx ⇒ − cos x = x sin x − ∫ x cos x dx ⇒ ∫ x cos x dx = x sin x + cos x + C

25. (i) Let, I = ∫ e sin bx dx

ax

e ax  e ax 
= sin bx. − ∫ (b cos bx )  dx [Integrating by parts]
a  a 
e ax sin bx b
= − ∫ cos bx e ax dx
a a
e ax sin bx b  e ax  e ax  
= −  cos bx + ∫ b sin bx  dx  [Again integrating by parts]
a a a  a  

e ax sin bx b cos bx e ax b 2
∴ I= − − 2 ∫ sin bx eax dx
a a2 a
2
 a sin bx − b cos bx  b  b2   a sin bx − b cos bx 
⇒ I = eax  −
 a2 I ⇒  1 + 2  I = e ax  
 a2   a   a2 
e ax
Hence, I = [ a sin bx − b cos bx ] + C
a2 + b2
e ax  eax 
(ii) Let, I = ∫ e ax cos bx dx = cos bx. − ∫ b( − sin bx).  dx [Integrating by parts]
a  a 

e ax cos bx b
= + ∫ sin bx e ax dx
a a
e ax cos bx b  e ax  e ax  
= + sin bx − ∫ b cos bx  dx  [Again integrating by parts]
a a a  a  

e ax cos bx b b2
= + 2 sin bx. e ax − 2 ∫ e ax cos bx dx
a a a
e ax cos bx b b2  b2  ae ax cos bx + be ax sin bx
∴I = + 2 sin bx. e ax − 2 .I ⇒ 1 + 2  I =
a a a  a  a2

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(BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES 17

e ax
Hence. I = [a cos bx + b sin bx] + C
a2 + b2
(iii) Let, I = sin ( log x ) dx = ∫ sin ( log x )1.dx
1
= sin ( log x ) .x − ∫ cos ( log x ) . .xdx [Integrating by parts]
x
= x sin ( log x ) − ∫ cos ( log x ) .1dx

⇒ I = x sin ( log x ) − x cos ( log x ) − ∫ sin ( log x ) dx ⇒ I = x sin ( log x ) − x cos ( log x ) − I

⇒ I + I = x sin ( log x ) − cos ( log x )  + C ⇒ 2 I = x sin ( log x ) − cos(( log x )  + C

x
Hence , I = sin ( log x ) − cos ( log x )  + C
2
1
(iv) Let, I = ∫ e− x cos 2 x cos 4 x dx = ∫ e − x [ 2 cos 4 x cos 2 x ] dx
2
1
= ∫ e− x [ cos 6 x + cos 2 x ] dx [∵ 2 cos A cos B = cos ( A + B ) + cos ( A − B )
2
1
∴ I =  ∫ e − x cos 6 x dx + ∫ e − x cos 2 x dx  …(1)
2 
Let, I1 = ∫ e− x cos 6 x dx and I 2 = ∫ e− x cos 2 x dx

(sin 6 x)   sin 6 x  
Now, I1 = ∫ e− x cos 6 x dx = e− x
6
(
− ∫  −e− x  )
 6 
  dx

e− x sin 6 x 1 − x
⇒ I1 = + ∫ e sin 6 x dx
6 6
e− x sin 6 x 1  − x  − cos 6 x   − x  − cos 6 x   
⇒ I1 =
6
+ e 
6  6  
(
 − ∫  −e 
 6
)  dx 
 
e− x sin 6 x e − x cos 6 x 1 e− x sin 6 x e − x cos 6 x 1
⇒ I1 = − − ∫ e− x cos 6 x dx ⇒ I1 = − − I1
6 36 36 6 36 36
1 e− x 37 e− x
⇒ I1 + I1 = [ 6sin 6 x − cos 6 x ] ⇒ I1 = [ 6sin 6 x − cos 6 x ]
36 36 36 36
e− x
∴ I1 = [6 sin 6 x − cos 6 x ] + C1 …(2)
37
(sin 2 x)   sin 2 x  
And I 2 = ∫ e − x cos 2 x dx ⇒ I 2 = e− x
2
(
− ∫  −e− x  )
 2 
  dx

1 −x 1
⇒ I2 = e sin 2 x + ∫ e − x sin 2 x dx
2 2
1 1 − cos 2 x   − x  − cos 2 x  
⇒ I 2 = e − x sin 2 x + e − x   − ∫ ( − e )   dx 
2 2  2    2  
1 1  −e− x cos 2 x 1 − x 
⇒ I 2 = e− x sin 2 x +  − ∫ e cos 2 x dx 
2 2 2 2 
1 1 1
⇒ I 2 = e − x sin 2 x − e − x cos 2 x − ∫ e − x cos 2 x dx
2 4 4

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18 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES
1 1 1 1 1
⇒ I 2 = e − x sin 2 x − e − x cos 2 x − I 2 ⇒ I 2 + I 2 = e− x [ 2sin 2 x − cos 2 x ]
2 4 4 4 4
5 1 −x e− x
⇒ I 2 = e [ 2 sin 2 x − cos 2 x ] ⇒ I 2 = [ 2sin 2 x − cos 2 x ] + C2 …(3)
4 4 5
Now, from (1), (2) and (3) , we get
1  e− x e− x 
I=  ( 6 sin 6 x − cos 6 x ) + ( 2sin 2 x − cos 2 x )  + C , where C = C1 + C2
2  37 5 
e− x 1 1 
Hence, I =  37 (6sin 6 x − cos 6 x) + 5 (2sin 2 x − cos 2 x)  + C
2
26. (i) Let, I = ∫ sin 3 xdx
1 1
Put x=z ⇒ dx = dz ⇒ dx = dz ⇒ dx = 2 z dz
2 x 2z
3 1
∴ I = ∫ ( sin 3 z )2 z dz = 2 ∫ z sin 3 z dz = 2 ∫ z  sin z − sin 3z  dz
4 4 
 3 3 3 3 1 
∵ sin 3 A = 3sin A − 4sin A ⇒ 4sin A = 3sin A − sin 3 A ⇒ sin A = 4 sin A − 4 sin 3 A
3 1
= ∫ z sin z dz − ∫ z sin 3 z dz
2 2
Integrating by parts, we get,
3 d   1 d  
I=  z.∫ sin z dz − ∫  ( z ) .∫ sin z dz  dz  −  z.∫ sin 3z dz − ∫  ( z ) .∫ sin 3z dz  dz 
2  dz   2  dz  
3 1   − cos 3z   − cos 3z  
z. ( − cos z ) − ∫ 1. ( − cos z ) dz  −  z. 
3  ∫ 
=  − 1.   dz
2  2  3  
3 3 1 1
= − z cos z + ∫ cos z dz + z cos 3 z − ∫ cos 3 z dz
2 2 6 6
3 3 1 1  sin 3 z  3 3 1 1
= − z cos z + sin z + z cos 3 z −   + C = − z cos z + sin z + z cos 3 z − sin 3 z + C
2 2 6 6 3  2 2 6 18
3 3 1 1
=− x cos x + sin x + x cos 3 x − sin 3 x + C
2 2 6 18
(ii) Let, I = ∫ sec −1 x dx
1 1
Put x=z ⇒ dx = dz ⇒ dx = dz ⇒ dx = 2 z dz
2 x 2z
∴ I = ∫ ( sec −1 z ) 2 z dz = 2 ∫ z sec −1 z dz

 d  
Integrating by parts, we get, I = 2 sec −1 z.∫ z dz − ∫  ( sec −1 z ) .∫ zdz  dz 
  dz  
 z2 1 z2  z
= 2 sec −1 z. − ∫ . dz  = z 2 sec −1 z − ∫ dz
 2 z z2 − 1 2  z2 − 1
z
⇒ I = z 2 sec −1 z − I1 …(1) where I1 = ∫ dz
2
z −1

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 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES 19
1
Put z 2 − 1 = y ⇒ 2z dz = dy ⇒ zdz = dy
2
 − 1 +1 
1  1  1 −1/2 1  y 2  1 y1/ 2
∴ I1 = ∫  dy  = ∫ y dy =  1  = × = y
y 2  2 2  − + 1 2 (1/ 2 )
 2 
= z2 −1 ∵ y = z 2 − 1

Putting the value of I1 in equation (1), we get, I = z 2 sec−1 z − ( )

z 2 − 1 + C = x sec −1 x − x − 1 + C

(iii) Let, I = ∫ cos x.log sin x dx

d 
Integrating by parts, we get, I = log sin x.∫ cos x dx − ∫  ( log sin x ) .∫ cos x dx  dx
 dx 
 1 
= ( log sin x )( sin x ) − ∫  cos x  .sin x dx
 sin x 
= sin x ( log sin x ) − ∫ cos x dx = sin x ( log sin x ) − sin x + C
1
27. (i) Let, I = ∫ x5 cos x 3dx = ∫ x 3 cos x3 .x 2 dx . Put x3 = t ⇒ 3 x 2 dx = dt ⇒ x 2 dx = dt
3
1 1
∴ I = ∫ t cos t dt = t sin t − ∫ {1.sin t } dt  [Integrating by parts]
3 3 
1 1
= [t sin t + cos t ] + C . Hence, I =  x 3 sin x3 + cos x3  + C
3 3
(ii) Let, I = ∫ x tan x sec2 x dx . Put tan x = z ⇒ x = tan −1 z ⇒ sec2 xd x = dz

z2 1 z2
( )
∴ I = ∫ tan −1 z z dz = tan −1 z.
2
−∫ . dz
1+ z2 2
[Integrating by parts]

z2 1 z2 z2 1 z2 +1 −1
tan −1 z − ∫
2 ∫ 1+ z2
−1
= dz = tan z − dz
2 2 1+ z2 2
z2 1  1  z2 1
= tan −1 z − ∫ 1 − 2 
dz = tan −1 z −  z − tan −1 z  + C
2 2  1+ z  2 2
x tan 2 x 1 x
Hence, I = − tan x + + C
2 2 2
sin −1 x
(iii) Let, I = ∫ 3
dx . Put sin −1 x = t ⇒ x = sin t ⇒ dx = cos t dt
(1 − x ) 2 2

t cos t t cos t
∴ I =∫ 3
dt = ∫ 3
dt = ∫ t sec 2 t dt = t. ( tan t ) − ∫ {1. tan t}dt [Integrating by parts]
2 cos t
(1 − sin t ) 2

= t.(tan t ) + log | cos t | +C

 x 
(
= sin −1 x . ) x
+ log 1 − x 2 + C 2
∵ cos t = 1 − x and tan t = 
1− x 2  1 − x2 
x sin −1 x 1
Hence, I = + log |1 − x 2 | +C
1 − x2 2

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20 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES
−1
x tan x
(iv) Let, I = ∫ 3
dx . Put tan −1 x = t ⇒ x = tan t ⇒ dx = sec2 t dt
(1 + x ) 2 2

∴ I =∫
( tan t ) t .sec2 t dt
3

(1 + tan t ) 2 2

( tan t ) t dt =
=∫
sec t ∫ t sin t dt = t ( − cos t ) − ∫ 1. ( − cos t ) dt [Integrating by parts]

= −t cos t + sin t + C
− tan −1 x x  x 1 
Hence, I = + +C ∵ sin t = and cos t = 
1 + x2 1 + x2  1 + x2 1 + x2 
cos x + sin x cos x + sin x
28. (i) Let I = ∫ cos 2 x.log dx ∴ I = ∫ log .cos 2 xdx
cos x − sin x cos x − sin x
Integrating by parts, we get,
cos x + sin x sin 2 x  − sin x + cos x − sin x − cos x  sin 2 x
I = log . −∫ − . dx
cos x − sin x 2  cos x + sin x cos x − sin x  2
2 2

= sin x cos x log

cos x + sin x
−∫
( cos x − sin x ) + ( cos x + sin x ) . sin 2 x dx
cos x − sin x cos 2 x − sin 2 x 2
cos x + sin x 2 sin 2 x cos x + sin x 1
= sin x cos x log −∫ . dx = sin x cos x log − log | sec 2 x | +C
cos x − sin x cos 2 x 2 cos x − sin x 2
(ii) Let I = ∫ x tan −1 ( 2 x + 3) dx ∴ I = ∫ tan −1 ( 2 x + 3) x dx
2
2 x2
Integrating by parts, we get, I = tan −1 ( 2 x + 3) . x − ∫ 2
. dx
2 1 + ( 2 x + 3) 2

x2 x2 x2 x2
= tan ( 2 x + 3 ) − ∫
−1
2
dx = tan −1
( 2 x + 3 ) − ∫ 4 x 2 + 12 x + 10 dx
2 1 + ( 2 x + 3) 2

x2 1 x2 x2 1
= tan −1 ( 2 x + 3) − ∫ 2 dx ∴ I= tan −1 ( 2 x + 3) − I 1 (say) …(i)
2 4 x + 3x + 5 / 2 2 4
x2  3x + 5 / 2  x +5/6
Now, I1 = ∫ 2
dx = ∫ 1 − 2 dx = x − 3∫ 2 dx
x + 3x + 5 / 2  x + 3x + 5 / 2  x + 3x + 5 / 2
3 2x + 5 / 3 3 ( 2 x + 3) + ( 5 / 3 − 3)
= x− ∫ dx = x − ∫ dx
2
2 x + 3x + 5 / 2 2 x 2 + 3x + 5 / 2
3 2x + 3 dx 3 5
= x− ∫ 2 dx + 2 ∫ 2 = x − log x 2 + 3x + + 2 ∫
dx
2 x + 3x + 5 / 2 x + 3x + 5 / 2 2 2 
2
3 1
x+  +
 2 4
3 5 dz 3
= x − log x 2 + 3 x + + 2 ∫ 2 2
, where z = x + ∴ dz = dx
2 2 z + (1/ 2 ) 2

3 5 2 z 3 5   3 
= x − log x 2 + 3 x + + tan −1 = x − log x 2 + 3 x + + 4 tan −1 2  x +  
2 2 1/ 2 1/ 2 2 2   2 
3 5
∴ I1 = x − log x 2 + 3 x + + 4 tan −1 ( 2 x + 3)
2 2

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 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES 21

x2 1 3 5 
∴ From (i), I = tan −1 ( 2 x + 3) −  x − log x 2 + 3 x + + 4 tan −1 ( 2 x + 3)  + C
2 4 2 2 
 x2  x 3 5
Hence, I =  − 1 tan −1 ( 2 x + 3) − + log x 2 + 3x + + C
 2  4 8 2

( )
2
29. (i) Let, I = ∫ sin −1 x dx . Put sin −1 x = t ⇒ x = sin t ⇒ dx = cos t dt

∴ I = ∫ t 2 cos t dt = t 2 sin t − ∫ 2t sin t dt [Integrating by parts]

= t 2 sin t − 2 t ( − cos t ) + ∫ 1. ( cos t ) dt  = t 2 sin t + 2t cos t − 2sin t + C

 

( )
2
Hence, I = x sin −1 x + 2 sin −1 x 1 − x 2 − 2 x + C

(ii) Let, I = ∫ cos −1 xdx . Put cos −1 x = z ⇒ x = cos z ⇒ x = cos 2 z

⇒ dx = −2 cos z sin z dz ⇒ dx = − sin 2 z dz
  − cos 2 z   − cos 2 z  
∴ I = − ∫ z sin 2 z dz = −  z   − ∫ 1.   dz  [Integrating by parts]
  2   2  
 − z cos 2 z 1   − z cos 2 z 1 sin 2 z  z cos 2 z sin 2 z
= − + ∫ cos 2 z dz  = −  + .  +C = − +C
 2 2   2 2 2  2 4
z ( 2 cos 2 z − 1) sin z cos z z ( 2 cos 2 z − 1) 1 − cos 2 z .cos z
= − +C = − +C
2 2 2 2

=
( cos −1
x ( 2 x − 1) )− x − x2
+C Hence, I =
( 2 x − 1) cos −1 x
−
x − x2
+C
2 2 2 2
π 
−1 −1 sin −1 x −  − sin −1 x 
sin x − cos x  2 dx  π
(iii) Let, I = ∫ dx = ∫ ∵ sin
−1
x + cos −1 x = 
sin −1
x + cos −1 x π 2
2
2  π 4
∫ x −  dx = ∫ sin −1 x dx − ∫ 1.dx
−1
=  2sin
π  2 π
4 −1
∴ I = ∫ sin x dx − x + C …(1)
π
4
∫ sin
−1
Let I1 = x dx . Put sin −1 x = θ ⇒ x = sin θ
π
⇒ x = sin 2 θ ⇒ dx = 2sin θ cos θ dθ ⇒ dx = sin 2θ dθ
4 4  −θ cos 2θ 1  4  −θ cos 2θ 1 
π∫
∴ I1 = θ sin 2θ dθ =  + ∫ cos 2θ dθ  =  + sin 2θ 
π 2 2  π 2 4 
4  −θ 1 
= 
π 2
( 2
)
× 1 − 2 sin 2 θ + sin θ 1 − sin 2 θ 

−2 2
Hence, I1 = sin −1 x (1 − 2 x ) + x 1− x
π π
2 2
On putting the value of I1 in (1), we get, I = ( 2 x − 1) sin −1 x+ x − x2 − x + C
π π
30. (i) Let, I = ∫ e ( cot x + log x sin x ) dx . Here, f ( x ) = log sin x
x
⇒ f ′ ( x ) = cot x

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22 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES
∴ I = e log sin x + C
x

Alternatively : Here, f ( x ) = log sin x ⇒ f ′ ( x ) = cot x

∴ ∫ e x [ cot x + log x sin x ] dx = ∫ e x log ( sin x ) dx + ∫ e x cot xdx

= log ( sin x ) .∫ e x dx − ∫ cot x.e x dx  + ∫ e x cot xdx [Integrating by parts]

 
= e x log ( sin x ) − ∫ e x cot x dx + ∫ e x cot x dx . Hence, I = e x log sin x + C .

1 1  1 ex
(ii) Let, I = ∫ e x  − 2  dx = ∫ e x . dx − ∫ 2 dx
x x  x x
1  1    ex
=  ∫ e x dx − ∫  − 2  e x dx  dx  − ∫ 2 dx [Integrating by parts]
x  x    x
ex ex ex ex
= + ∫ 2 dx − ∫ 2 dx . Hence, I = + C .
x x x x

(iii) Let, I = ∫
xe x
dx = ∫
( x + 1 − 1) e x dx =  ( x + 1) − 1  e x dx
( x + 1)
2
( x + 1)
2 ∫  ( x + 1)2 ( x + 1)2 
 
 1 1  1 ex
= ∫ ex  − 2  dx = ∫ x + 1 ∫ ( x + 1)2 dx
e x
. dx −
 x + 1 ( x + 1) 
 1  −1    ex
= 
 x +1 ∫ ∫  ( x + 1)2  ∫   ∫ ( x + 1)2 dx
e x
dx −   e x
dx dx − [Integrating by parts]
    
ex ex ex ex
= +∫ 2
dx − ∫ ( x + 1)2 dx . Hence, I = +C
x +1 ( x + 1) x +1

 1 − sin x   1 sin x 
(iv) Let, I = ∫ e x   dx = ∫ e x  − dx
1 − cos x  1 − cos x 1 − cos x 
 x x
 2sin cos 
1 2 2 dx = e x  1 cosec2 x − cot x  dx = e cosec 2 x dx − e x cot x dx
x
= ∫ ex  − ∫  2 ∫2 ∫
 2sin 2 x x  2 2  2 2
2sin 2
 2 2 
ex x  x  1 x  
=∫ cosec 2 dx − cot ∫ e x dx − ∫ − cosec 2 ∫ e x dx  dx  [Integrating by parts]
2 2  2  2 2  
1 x x x 1 x x
=
2 ∫ e cosec2 dx − e x cot − ∫ e x cosec2 dx . Hence, I = −e x cot + C .
2 2 2 2 2
(v) Let, I = ∫ e ( tan x + log sec x ) dx = ∫ e log sec xdx + ∫ e tan xdx
x x x

 {  }
= log sec x ∫ e x dx − ∫ tan x.∫ e x dx dx  + ∫ e x tan xdx [Integrating by parts]

= e x log sec x − ∫ e x tan xdx + ∫ e x tan x dx . Hence, I = e x log sec x + C .

 1 1  1 1
(vi) Let, I = ∫  − dx = ∫ .1dx − ∫ dx
 log x ( log x )2  log x ( log x )
2
 
 1  −1  1   1
= ∫ 1.dx − ∫   . .x  dx  − ∫ dx [Integrating by parts]
 log x ( log )
2
 x  ( log )
2
  x    x

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 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES 23
x 1 1 x
= +∫ 2
dx − ∫ 2
dx . Hence, I = +C .
log x ( log x ) ( log x ) log x
(vi) Let, I = ∫ {sin ( log x ) + cos ( log x )}dx = ∫ sin ( log x ) dx + ∫ cos ( log x ) dx
  1  
= sin ( log x ) ∫ 1.dx − ∫ cos ( log x ) . ∫ 1.dx  dx  + ∫ cos ( log x ) dx
  x  
= x sin ( log x ) − ∫ cos ( log x ) dx + ∫ cos ( log x ) dx . Hence, I = x sin ( log x ) + C .
 1  1
(vii) Let, I = ∫ log ( log x ) + 2 
dx = ∫ log ( log x ) dx + ∫ 2
dx
 ( log x )  ( log x )
  1 1   1
= log ( log x ) .∫ 1.dx − ∫  . ∫ 1.dx  dx  + ∫ 2
dx [Integrating by parts]
  log x x   ( log x )
1 1
= log ( log x ) .x − ∫ dx + ∫ 2
dx
log x ( log x )
 1  −1 1   1
= x log ( log x ) −  .∫ 1.dx − ∫  2
. ∫ 1. dx  dx  + ∫ 2
dx
 log x  ( log x ) x   ( log x )
[Again Integrating by parts]
x 1 1
= x log ( log x ) − −∫ 2
dx + ∫ 2
dx
log x ( log x ) ( log x )
x
Hence, I = x log ( log x ) − +C
log x

I = ∫ e− x tan −1 ( e x ) dx
dt dt
31. Put e x = t ⇒ e x dx = dt ⇒ dx x
⇒ dx =
e t
1 dt
∴ I = ∫ ⋅ tan −1 t ⋅
t t
1
∫ t 2 tan tdt
−1

 1 1 1
= ( tan −1 t )  −  + ∫ 2
dt
 t  1+ t t
tan −1 t t 1
=− +∫ 2 2 dt Put t 2 = y ⇒ 2tdt = dy ⇒ tdt = dy
t t ( t + 1) 2

tan −1 t 1 dy tan −1 t 1  1 1 
=− + ∫ =− + ∫ − dy
t 2 y ( y + 1) t 2  y y +1 
tan −1 t 1 tan −1 t 1
=− + log y − log ( y + 1)  = − + log t 2 − log ( t 2 + 1) 
t 2 t 2
tan −1 t 1 tan −1 e x 1
=− + log t − log ( t 2 + 1) = − + log e x − log ( e 2 x + 1)
t 2 e x
2
1
= −e− x tan −1 e x + x − log ( e 2 x + 1)
2
ALTERNATIVE
Integrating by parts taking tan −1 ( e x ) as u , we get,

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24 (BOARD LEVEL, XII) BY R. K. MALIK’S NEWTON CLASSES
ex
( ) (
I = −e− x tan −1 e x − ∫ −e − x ) dx = −e − x tan −1 ( e x ) + ∫
dx
…(1)
1 + (ex )
2x 2
1+ e

dx 1 sec 2 θ
Now, ∫ 1+ =∫ . dθ [Putting e x = tan θ so that e x dx = sec 2 θ dθ ]
(e )
x 2 1 + tan 2 θ tan θ

 ex  1 1
= ∫ cot θ dθ = log sin θ = log 
2x
 ( ) (
= log e x − log 1 + e 2 x = x − log 1 + e 2 x
2 2
) ( )
 1+ e 
1
( )
∴ From (1), we get, I = −e − x tan −1 e x + x − log 1 + e2 x + C
2
( )
  x x  x x
32. Let, I = ∫ log  2 cos 2  − x tan  dx = ∫  log 2 + 2 log cos − x tan  dx
  2 2  2 2
x x
= x log 2 − ∫ x tan dx + 2 ∫ 1.log cos dx
2 2
  x 
 x  − sin  
2 1
= x log 2 − ∫ x tan dx + 2  x log cos − ∫ 
x x
.  dx
2  2 cos
x 2
 2 
x x x x
= x log 2 − ∫ x tan dx + 2 x log cos + ∫ x tan dx . Hence, I = x log 2 + 2 x log cos + C .
2 2 2 2
x log x x
33. Let, I = ( x log x − x ) sin −1 x − ∫ dx + ∫ dx …(1)
2
1− x 1 − x2
[Integrating by parts taking sin −1 x as u ]
x log x
Now in order to evaluate ∫ 1 − x2
dx . Put x = sin θ ⇒ dx = cos θ dθ

x log x
∴ ∫ 1 − x2
dx = ∫ sin θ log sinθ dθ = − cos θ log sin θ − ∫ − cos θ .cot θ dθ

cos 2 θ 1 − sin 2 θ
= − cos θ log sin θ + ∫ dθ = − cos log sin θ + ∫ dθ
sin θ sin θ
= − cos θ log sin θ + ∫ ( cosec θ − sin θ ) dθ = − cos θ log sin θ + ∫ cosec θ dθ − ∫ sin θ dθ
= − cos θ log sin θ + log cosec θ − cot θ + cos θ
 1 − 1 − x2 
= − 1 − x 2 log x + log   + 1 − x2 …(2)
 x 
 
x
Again, ∫ dx = − 1 − x 2 …(3)
2
1− x
[On putting 1 − x 2 = t ⇒ − 2x dx = dt ]
∴ From (1), (2) and (3), we get,
 1 − 1 − x2 
I = ( x log x − x ) sin x + 1 − x log x − log 
−1 2
− 1 − x2 − 1 − x2 + C
 x 
 
 1 − 1 − x2 
Hence, I = ( x log x − x ) sin −1 x + 1 − x 2 log x − log   − 2 1 − x2 + C
 x 
 

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