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INTEGRATION BY PARTS

Theorem: If f(x) and g(x) are two integrable functions then

∫ f (x) ⋅ g(x)dx = f (x) ∫ g(x)dx − ∫ f ′(x)  ∫ g(x)dx  dx .

Proof:
d  d d
f (x) ⋅ ∫ g(x)dx  = f (x)  ∫ g(x)dx  + ∫ g(x)dx ⋅ [ f (x)]
dx   dx   dx

= f (x)g(x) +  ∫ g(x)dx  f ′(x)

∴ ∫ f (x)g(x) + f ′(x) ∫ g(x)dx  dx = f (x) ∫ g(x) dx

⇒ ∫ f (x)g(x)dx + ∫ f ′(x)  ∫ g(x)dx  dx = f (x) ∫ g(x) dx

 

∴ ∫ f (x)g(x)dx = f (x) ∫ g(x) dx − ∫ f ′(x)  ∫ g(x)dx  dx

 

Note 1: If u and v are two functions of x then ∫ u dv = uv − ∫ v du .

Note 2: If u and v are two functions of x; u′, u″, u″′ …… denote the successive derivatives of u and
v1, v2, v3, v4, v5 … the successive integrals of v then the extension of integration by pairs is

∫ uvdx = uv1 − u′v2 + u′′v3 − u′′′v4 + ... .

Note 3: In integration by parts, the first function will be taken as the following order.
Inverse functions, Logarithmic functions, Algebraic functions, Trigonometric functions and
Exponential functions. (To remember this a phrase ILATE).

eax
∫ e cos bx dx = (a cos bx + b sin bx) + c
ax
Theorem:
a 2 + b2

Proof: Let I = ∫ eax cos bx dx = cos bx ∫ eax dx − ∫ d(cos bx)∫ eax dx  dx

eax eax
= cos bx − ∫ (− b sin bx) dx
a a
eax b
= cos bx + ∫ eax sin bx dx
a a

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eax b eax eax 

= cos bx + sin bx − ∫ b cos bx dx 
a a a a 

eax b b2
= cos bx + 2 eax sin bx − 2 I
a a a

 b 2  1 ax
⇒ I 1 + 2  = 2 e [ a cos bx + b sin bx ]
 a  a
 a 2 + b 2  1 ax
⇒ I 2  = 2 e [ a cos bx + b sin bx ]
 a  a

eax
∴I = 2 [ a cos bx + b sin bx ] + c
a + b2

eax
Theorem: ∫ eax sin bx dx = (a sin bx − b cos bx)
a 2 + b2

Proof : Let I = ∫ eax sin bx dx = sin bx ∫ eax dx − ∫ d(sin bx) ∫ eax dx  dx

eax eax
= sin bx − ∫ b cos bx dx
a a
1 b
= eax sin bx − ∫ eax cos bx dx
a a

1 b eax eax 
= eax sin bx −  cos bx − ∫ (−b sin bx) dx 
a a a a 

1 b b2
= eax sin bx − 2 eax cos bx − 2 ∫ eax sin bx dx
a a a
1 ax b ax b2
= e sin bx − 2 e cos bx − 2 I
a a a

 a2  1
⇒ I 1 + 2  = 2 eax [ a sin bx − b cos bx ]
 b  a
 a 2 + b 2  eax
⇒ I 2  = 2 [ a sin bx − b cos bx ]
 b  a

eax
∴I = [ a sin bx − b cos bx ] + c
a 2 + b2

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∫ e [f (x) + f ′(x)] dx = e f (x) + c

x x
Theorem:

Proof:

∫ e [f (x) + f ′(x)] dx = ∫ e f (x)dx + ∫ e f ′(x)dx

x x x

= f (x) ∫ e x dx − ∫ d[f (x)]∫ e x dx  dx + ∫ e x f ′(x)dx

 
= f (x)e x − ∫ f ′(x)e x dx + ∫ e x f ′(x)dx = e x f (x) + c

Note: ∫ e− x [ f (x) − f ′(x)] dx = −e− x f (x) + c

Very Short Answer Questions

Evaluate the following integrals.

 (2n + 1)π 
1. ∫ x sec
2
x dx on I⊂R\ : n is an integer 
 2 

Sol. ∫ x sec 2 x dx = x(tan x) − ∫ tan x dx

= x tan x − log | sec x | +C

 −1 1 
∫e  tan x +  dx , x ∈ R.
x
2.
 1+ x2 

Sol.
1
Let f (x) = tan −1 x so that f ′(x) =
1+ x2


∴ ∫ e x  tan −1 x +

1 
1+ x 2
x −1 x
(
 dx = e tan x + C ∵ ∫ e [ f (x) + f ′(x) ] dx = e ⋅ f (x) + C
x
)
log x
3. ∫ dx on (0, ∞).
x2

log x  1 1 1
Sol. ∫ 2
dx = (log x)  −  + ∫ ⋅ dx
x  x x x

1 1
= − log x − + C
x x

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4. ∫ (log x)
2
dx on (0, ∞).

1
Sol. ∫ (log x) 2 dx = (log x) 2 x − ∫ x ⋅ 2 log x ⋅ dx
x

= x(log x)2 − 2 ∫ log x dx

 1 
= x(log x)2 − 2  x log x − ∫ x dx 
 x 
= x(log x)2 − 2x ⋅ log x + x + c

 π 
5. ∫e
x
(sec x + sec x tan x)dx on I ⊂ R \ (2n + 1) : n ∈ Z 
 2 

Sol. ∫ e x (sec x + sec x tan x)dx = e x ⋅ sec x + C

(∵ ∫ e [f (x) + f ′(x)]dx = e f (x) + C )

x x

∫e
x
6. cos x dx on R.

Sol. I = ∫ e x cos x dx = e x sin x − ∫ sin x ⋅ e x dx

= e x ⋅ sin x + e x ⋅ cos x − ∫ e x ⋅ cos x dx

= e x (sin x + cos x) − I

2I = e x (sin x + cos x)
ex
I= (sin x + cos x) + C
2

∫e (sin x + cos x)dx on R.

x
7.

Sol. ∫ e x (sin x + cos x)dx

f (x) = sin x ⇒ f ′(x) = cos x

∴ ∫ e x (sin x + cos x)dx = e x ⋅ sin x + C

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 1  1 
8. ∫ + x
on   2n −  π,  2n + π n∈Z
2  
(tan x log sec x)e dx
 2 

1
Sol. let f = log | sec x |⇒ f ' = ⋅ sec x ⋅ tan x ⋅
sec x

= tan x

∫ (tan x + log sec x)e

x
(
dx = e x ⋅ log | sec x | + C ∵ ∫ e x [f (x) + f ′(x)]dx = e x f (x) + C )

Short Answer Questions

Evaluate the following integrals.

1. ∫x
n
log x dx on (0, ∞), n is a real number and n ≠ –1.

x n +1 1 1
Sol. ∫ x n log x dx = (log x) − ∫ x n +1 dx
n +1 n +1 x

x n +1 (log x) 1
=
n +1
−
n +1 ∫ x n dx

x n +1 (log x) x n +1
= − +C
n +1 (n + 1) 2

x n +1
= [(n + 1) log x − 1] + C
(n + 1) 2

∫ log(1 + x
2
2. )dx on R.

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

1
= log(1 + x 2 ) ⋅ x − ∫ x 2x dx
1 + x2
1 + x2 −1
= x log(1 + x 2 ) − 2 ∫ dx
1+ x2

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dx
= x log(1 + x 2 ) − 2 ∫ dx + 2 ∫
1+ x2
= x log(1 + x 2 ) − 2x + 2 tan −1 x + C

3. ∫ x log x dx on (0, ∞).

Sol. ∫ x log x dx =

2 2 1
= log x ⋅ x 3/ 2 − ∫ x 3/ 2 ⋅ dx
3 3 x

2 3/ 2 2
= x (log x) − ∫ x1/ 2dx
3 3

2 3/ 2 2 x 3/ 2
= x (log x) − +C
3 3 3/ 2

2 3/ 2 4
= x log x − x 3/ 2 + C
3 9

4. ∫e
x
dx on (0, ∞).

Sol. let x = t ⇒ x = t 2 , dx = 2t dt

∫e
x
dx = 2 ∫ t e t dt = 2  te t − ∫ e t dt 
 
= 2(te t − e t ) + C
= 2 xe x
− 2e x
+C

∫x
2
5. cos x dx on R.

Sol. ∫ x 2 cos x dx = x 2 (sin x) − ∫ sin x(2x dx)

= x 2 sin x + 2 ∫ x(− sin x)dx

= x 2 ⋅ sin x + 2[x cos x − ∫ cos x dx]
= x 2 sin x + 2x cos x − 2sin x + c

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∫ x sin
2
6. x dx on R.

1
Sol. ∫ x sin 2 x dx = ∫ x(1 − cos x)dx
2

1
=  ∫ xdx − ∫ x cos 2x dx 
2
1  x 2  sin 2x 1 
=  − x ⋅ − ∫ sin 2x dx 
2  2  2 2 

x2 x 1
= − sin 2x + ∫ sin 2x dx
4 4 4
x2 x 1
= − sin 2x − cos 2x + C
4 4 8

∫ x cos
2
7. x dx on R.

1
Sol. ∫ x cos2 x dx = ∫ x(1 + cos 2x)dx
2

1
=  ∫ x dx + ∫ x cos 2x dx 
2
1  x 2  sin 2x 1 
=  + x − ∫ sin 2x dx 
2  2  2 2 

x2 x 1
= + sin 2x − ∫ sin 2x dx
4 4 4
x2 x 1
= + sin 2x + cos 2x + C
4 4 8

8. ∫ cos x dx on R.

Sol. x = t 2 ⇒ dx = 2t dt

I = 2∫ t ⋅ cos t dt = 2(t sin t − ∫ sin t dt)

= 2(t sin t + cos t) + C

= 2 x sin x + 2 cos x + C

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 π 
9. ∫ x sec
2
2x dx on I ⊂ R \ (2nπ + 1) : n ∈ Z 
 4 

tan 2x 1
Sol. ∫ x sec2 2x dx = x − ∫ tan 2x dx
2 2

tan 2x 1 1
=x − ⋅ log | sec 2x | + C
2 2 2
tan 2x 1
=x − log | sec 2x | + C
2 4

10. ∫ x cot
2
x dx on I ⊂ R \{nπ : n ∈ Z} .

Sol. ∫ x cot 2 x dx = ∫ x(csc 2 x − 1)dx

= ∫ x csc2 x dx − ∫ x dx

x2
= x(− cot x) + ∫ cot x dx −
2
x2
= − x cot x + log | sin x | − +C
2

 π 
11. ∫e
x
(tan x + sec 2 x)dx on I ⊂ R \ (2n + 1) : n ∈ Z 
 2 

Sol. f (x) = tan x ⇒ f ′(x) = sec 2 x dx

I = ∫ e x [ f (x) + f ′(x) ]dx = e x f (x) + C

= e x tan x + C

 1 + x log x 
∫e  dx on (0, ∞).
x
12. 
 x 

 1 + x log x  x 1
Sol. ∫ e x   dx = ∫ e  log x +  dx
 x   x

= e x log x + C

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dx
13. ∫ (x 2 + a 2 )2 ,(a > 0) on R.

Sol: Take substitution x = a tan θ

So that dx = a sec2θ dθ

dx a sec2 θdθ
∴∫ = ∫ (a 2 tan 2 θ + a 2 )2
(x 2 + a 2 ) 2

a sec2 θdθ 1 sec2 θdθ

=∫ = ∫
a 4 (1 + tan 2 θ)2 a 3 sec4 θ

1
= 3 ∫
cos 2 θdθ
a

1  1 + cos 2θ 
= ∫  2  dθ
a3 

1
=  ∫ 1 ⋅ dθ + ∫ cos 2θ dθ 
2a 3 

1  1 
= θ + 2 sin 2θ
2a 3

1  −1  x  1  −1  x   
=  tan   + sin  2 tan     + c
2a 3  a 2   a  

1 x 1   x 
= 3
tan −1   + 3 sin  2 tan −1    + c.
2a  a  4a   a 

14. ∫e
x
log(e2x + 5e x + 6)dx on r.

Sol: e2x + 5ex + 6 = (e x )2 + 5e x + 6

= (e x )2 + 3e x + 2e x + 6
= e x (e x + 3) + 2(e x + 3)
= (e x + 3)(e x + 2)

∫e
x
log(e2x + 5e x + 6)dx

= ∫ e x log[(e x + 2)(e x + 3)]dx

= ∫ e x log(e x + 2)dx + ∫ e x log(e x + 3) dx (∵ log ab = log a + log b)

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Let ex = t then ex dx = dt

∴ ∫ e x log(e2x + 5e x + 6)dx

= ∫ log(t + 2)dt + ∫ log(t + 3)dt

t t
= log(t + 2)t − ∫ dt + log(t + 3) ⋅ t − ∫ dt
t+2 t +3

(using integration by parts on two integrals)

 (t + 2) − 2   (t + 3) − 3 
= t ⋅ log(t + 2) − ∫   dt + t ⋅ log(t + 3) − ∫  dt
 t+2   t +3 

dt dt
= t ⋅ log(t + 2) − ∫ dt + 2 ∫ + t log(t + 3) − ∫ dt + 3∫
t+2 t +3
= t log(t + 2) − t + 2 log(t + 2) + t log(t + 3) − t + 3log(t + 3)
= 2 log | (t + 2) | +3log | (t + 3) | −2t + t[log(t + 2)(t + 3)]

= t[log(t 2 + 5t + 6)] − 2t + 2log | t + 2 | +3log | t + 3 | +c

= ex [log(e2x + 5ex + 6)] − 2e x + 2log | ex + 2 | +3log | ex + 3 | +c.

15. ∫ cos(log x)dx on (0, ∞).

Sol: Let I = ∫ cos(log x)dx = ∫ cos(log x)1 ⋅ dx

Take u = cos(log x) and v = 1 and using integration by parts successively.

1
I = cos(log x)x − ∫ − sin(log x) x ⋅ dx
x

= x cos(log x) + ∫ sin(log x)dx

1
= x cos(log x) + sin(log x) ⋅ x − ∫ cos(log x) ⋅ x ⋅ dx
x
= x cos(log x) + x ⋅ sin(log x) − ∫ cos(log x)dx
= x[cos(log x) + sin(log x)] − 1

∴ 2I = x[cos(log x) + sin(log x)]

x
⇒ I = [cos(log x) + sin(log x)] + c
2

∴ ∫ cos(log x)dx =
x
[cos(log x) + sin(log x)] + c
2
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x+2
16. ∫e
x
dx on I ⊂ R \ {–3}
(x + 3) 2

x+2
Sol. ∫ e x dx
(x + 3) 2

Hint: ∫ e x [ f (x) + f ′(x) ] dx = e x − f (x) + C

 x + 3 − 1 
= ∫ ex  2
dx
 (x + 3) 

 1 (−1)  x 1 
= ∫ ex  +  dx = e  +C
 x + 3 (x + 3)   x +3
2

xe x
17. ∫ (x + 1)2 dx on I ⊂ R \ {–1}

xe x  x + 1 −1 x
Sol. ∫ dx = ∫  (x + 1)2  e dx
(x + 1)2  

 1 1  x
= ∫ − 2
e dx
 x + 1 (x + 1) 
 1  ( −1)  x
= ∫  + 2
e dx
 x + 1  (x + 1) 

Hint: ∫ e x [ f (x) + f ′(x) ] dx = e x − f (x) + C

 1  x ex
=  e + C = x +1 + C
 x +1 

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Long Answer Questions

Evaluate the following integrals.

−1
1. ∫ x tan x dx, x ∈ R

x2 1 2 1
Sol. ∫ x tan −1 x dx = (tan −1 x) − ∫x ⋅ dx
2 2 1+ x2

x 2 (tan −1 x) 1  1 
= − ∫ 1 −  dx
2 2  1+ x2 

x 2 (tan −1 x) 1
= − (x − tan −1 x) + C
2 2

x 2 (tan −1 x) x tan −1 x
= − + +C
2 2 2
(x 2 + 1) x
= tan −1 x − + C
2 2

2. ∫x
2
tan −1 x dx, x ∈ R .

x3 1 3 1
Sol. ∫ x 2 tan −1 x dx = (tan −1 x)
3 3 ∫ 1+ x2
− x dx

x 3 (tan −1 x) 1 x(x 2 + 1) − x
= − ∫ dx
3 3 1 + x2
x 3 (tan −1 x) 1 1 xdx
= − ∫ xdx + ∫
3 3 3 1+ x2

x 3 (tan −1 x) x 2 1
= − + log |1 + x 2 | +C
3 6 6

tan −1 x
3. ∫ x2
dx, x ∈ I ⊂ R \ {0}

tan −1 x 1  1 1 1
Sol. ∫ dx = ∫ tan −1 x = (tan −1
x)  −  + ∫ dx
x2 x2  x x 1+ x2

tan −1 x 1 2x dx
=− + ∫ 2
x 2 x (1 + x 2 )
tan −1 x 1  1 1 
=− + ∫ 2 −  (2x dx)
x 2  x 1+ x2 
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tan −1 x dx 1 2x dx
+∫
x 2 ∫ 1+ x2
= −
x

tan −1 x 1
=− + log | x | − log |1 + x 2 | +C
x 2

−1
4. ∫ x cos x dx, x ∈ (−1,1)

Sol. ∫ x cos −1 x

d 
= cos −1 ∫ x dx − ∫  [cos −1 x]∫ x dx  dx
 dx 

x2 −1 x 2
= cos −1 x − ∫ dx
2 1− x2 2
x2 1 x2
= cos −1 x + ∫ dx
2 2 1− x2

x2 −1 1 1 − x2 −1
= cos x − ∫ dx
2 2 1− x2

x2 1 1 1
= cos −1 x − ∫ 1 − x 2 dx + ∫ dx
2 2 2 1− x2

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

x2 1 1
= cos −1 x − x 1 − x 2 + sin −1 x + C
2 4 4

5. ∫x
2
sin −1 x dx, x ∈ (−1,1)

Sol. ∫ x 2 sin −1 x dx

x3 1 3  1 
= (sin −1 x)
3 3 ∫  1 − x 2
− x   dx



x 3 −1 1 x[1 − (1 − x 2 )]
= sin x − ∫ dx
3 3 1− x2

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x 3 −1 1 xdx 1
= sin x − ∫ + ∫ x 1 − x 2 dx
3 3 1− x2 3

x 3 −1 1 1 (1 − x 2 )3/ 2
= sin x + 1 − x 2 + +C
3 3 3 (3 / 2)(−2)

x 3 −1 1− x2 1
= sin x + − (1 − x 2 )3/ 2 + C
3 3 9

6. ∫ x log(1 + x)dx, x ∈ (−1, ∞)

Sol. ∫ x log(1 + x)dx

 x2  1 x2
= log(1 + x)   − ∫ dx
 2  2 1+ x
 
x2 1 1 − (1 − x 2 )
= log(1 + x) − ∫ dx
2 2 1+ x

x2 1 dx 1
log(1 + x) − ∫
2 1+ x 2 ∫
= + (1 − x)dx
2

x2 1 1 x2 
= log(1 + x) − log(1 + x) +  x −  + C
2 2 2  2 

(x 2 − 1) x x2
= log(1 + x) + − +C
2 2 4

7. ∫ sin x dx on (0, ∞).

Sol. put x = t 2 ⇒ dx = 2t dt

= 2  t(− cos t) + ∫ cos t dt 

= −2t cos t + 2 sin t

= −2 x cos x + 2 sin x + C

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8. ∫e
ax
sin(bx + c)dx , (a, b, c ∈ R, b ≠ 0) on R.

Sol.

Let I = ∫ eax sin(bx + c)dx

 cos(bx + c)  1
= eax  −  + ∫ cos(bx + c)e a dx
ax
 b  b

eax ⋅ cos(bx + c) a ax
=− + ∫ e cos(bx + c)dx
b b

eax ⋅ cos(bx + c) a  ax bx + c  1
b  b∫
=− +  e ⋅ sin − sin(bx + c)eax ⋅ a ⋅ dx
b b

eax ⋅ cos(bx + c) a ax a2
=− + 2 e sin(bx + c) − 2 I
b b b

 a2  eax a a 2 + b2 eax
1 + 2  I = − cos(bx + c) + 2 eax sin(bx + c) I = [a sin(bx + c) − b(cos(bx + c)]
 b  b b b2 b2

eax
∴I = [a sin(bx + c) − b(cos(bx + c)] + C1
a 2 + b2

9. ∫a
x
cos 2x dx on R(a > 0 and a ≠ 1).

Sol. ∫ a x cos 2x dx

sin 2x 1
= ax − ∫ sin 2x ⋅ a x log a dx
2 2

a x ⋅ sin 2x log a x
2 ∫
= + a (− sin 2x)dx
2

a x sin 2x log a x 2x 1
= + (a ⋅ cos − ∫ cos 2x ⋅ a x log a dx)
2 2 2 2

a x sin 2x a x log a cos 2x (log a)2

= + − I
2 4 4

 (log a)2  a x [2sin 2x + (log a) cos 2x]

1 +  I =
 4  4

4 + (log a) 2 a x [2sin 2x + (cos 2x) log a]

I=
4 4

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2 ⋅ a x ⋅ sin 2x + (a x ⋅ log a) cos 2x

∴I = +c
(log a) 2 + 4

−1  3x − x 
3
 1 1 
10. ∫ tan 
1 − 3x

2 
dx on I ⊂ R \ −
 3
,
3
.

 

Sol. Put x = tan t ⇒ dx = sec2 t dt

−1  3x − x 
3
Then ∫ tan 
  dx
 1 − 3x 
2

 3 tan t − tan 3 t  2
= ∫ tan −1 
 1 − 3 tan 2 t 
sec t dt
 

= ∫ tan −1 (tan 3t) sec 2 t dt = 3∫ t sec 2 t dt

 d  
= 3  t ∫ sec2 t dt − ∫  (t) ∫ sec2 t dt  dt 
  dt  
= 3[t(tan t) − ∫ (1) tan t dt]

= 3(t tan t − log | sec t |) + C

(
= 3 x ⋅ tan −1 x − log 1 + x 2 + C )
 3 
= 3x  tan −1 x − log(1 + x 2 )  + C
 2 
3
= 3x tan −1 (x) − log(1 + x 2 ) + C
2

−1
11. ∫ sinh x dx on R.

Sol. ∫ sinh −1 x dx = ∫ 1.sinh −1 x dx

1
= x.sinh −1 x − ∫ .xdx
1+ x 2

1 2x
= x.sinh −1 x − ∫ dx
= 2 1+ x2
1
= x.sinh −1 x − 2. 1 + x 2 + c
2
= x.sinh −1 x − 1 + x 2 + c

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−1
12. ∫ cosh xdx on [1, ∞].

Sol. ∫ cosh −1 xdx = ∫ 1.cosh −1 xdx

Apply integration of parts.

Ans. x cosh −1 x − x 2 − 1 + C

−1
13. ∫ tanh x dx on (–1, 1).

Sol. ∫ tanh −1 x dx = ∫ 1. tanh −1 x dx

= ∫ 1.tanh −1 x dx
1
= x.tanh −1 x − ∫ xdx
1− x2
1 −2x
= x.tanh −1 x + ∫ dx
2 1− x2
1
(
= x.tanh −1 x + log 1 − x 2 + c
2
)
14. Find ∫ eax cos(bx + c)dx on R where a, b, c are real numbers and b ≠ 0.

Sol. Let A = ∫ eax cos(bx + c)dx

Then from the formula for integration by parts

 sin(bx + c)   sin(bx + c) 
A = eax   − ∫ aeax   dx
 b   b

1 a
= eax sin(bx + c) − ∫ eax sin(bx + c)dx
b b

1 ax a   − cos(bx + c)  ax  cos(bx + c)  
= e sin(bx + c) − eax   − ∫ ae −  dx  + C1
b b  b   b  

1 ax a ax a2
= e sin(bx + c) + 2 e cos(bx + c) − 2 A + C 2
b b b

 a2  a ax 1 ax
1 + 2  A = 2 e cos(bx + c) + e sin(bx + c) + C2
 b  b b

(a 2 + b 2 )A = ae ax cos(bx + c) + beax sin(bx + c) + C3

eax
Hence A = [a cos(bx + c) + b sin(bx + c)] + K
a 2 + b2
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c3
Where k = a constant
a + b2
2

By taking c = 0, we get

eax
∫ e cos bx dx = [a cos bx + b sin bx] + K
ax
a 2 + b2

1− x
15. Evaluate ∫ tan −1 dx on (–1, 1).
1+ x

Sol. Put x = cosθ, θ∈ (0, π)dx = –sin θ dθ

1 − x 1 − cos θ 2 sin 2 θ / 2 θ
= = = tan 2
1 + x 1 + cos θ 2 cos θ / 2
2
2

−1 1− x θ
∫ tan dx = ∫ tan −1 tan 2 ( − sin θ)dθ
1+ x 2

 θ
= − ∫ tan −1  tan  (sin θ)dθ
 2
1
2∫
=− θ⋅ sin θ dθ

1
= − θ(− cos θ) − ∫ (− cos θ)dθ + C
2 

1
= (θ cos θ − sin θ) + C
2

=
1
2 (
x cos −1 x − 1 − x 2 + C )

 1 − sin x 
16. Evaluate ∫ e x   dx on I ⊂ R \ {2nπ
π : n ∈ Z}.
 1 − cos x 

1 − sin x 1 − sin x
Sol. =
1 − cos x 2 sin 2 x / 2

x x x x
1 − 2sin cos 2sin cos
2 2= 1 2 2
= −
2 x 2 x 2 x
2sin 2sin 2sin
2 2 2

1 x x
= csc2 − cot
2 2 2

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 1 − sin x  x1 2 x x
∫e  dx = ∫ e  csc − cot  dx
x

 1 − cos x  2 2 2

x
= ∫ e x [ f (x) + f ′(x)] dx where f (x) = − cot
2

x
= e x f (x) + C = −e x cot +C
2

 2x 
17. Evaluate ∫ tan −1  2 
dx on I ⊂ R \ (–1, 1).
 1− x 

Sol. Let x = tan θ⇒ dx = sec2θ dθ

2x 2 tan θ
= = tan 2θ
1− x 2
1 − tan 2 θ
 2x 
tan −1  2 
= tan −1 (tan 2θ) = 2θ + nπ
 1− x 

Where n = 0 if |x| < 1

= –1 if x > 1
= 1 if x < –1
1
We have dθ = dx and
1+ x2

1+x2 = 1 + tan2θ = sec2θ

 2x 
∴ ∫ tan −1  2 
dx
 1− x 

  2x   1
= ∫  tan −1  2 
(1 + x 2 ) dx
  1− x   1+ x2
= ∫ (2θ + nπ) ∫ sec2 θdθ

= 2 ∫ θ sec2 θdθ + nπ∫ sec2 θdθ + c

( )
= 2 θ tan θ − ∫ tan θdθ + nπ tan θ + c
= 2(θ tan θ + log | cos θ | + nπ tan θ + c

= (2θ + nπ) tan θ + 2log cos θ + c

= (2θ + nπ) tan θ + log cos 2 θ + c

= (2θ + nπ) tan θ + log sec2 θ + c

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 2x 
= x tan −1  2
− log(1 + x 2 ) + c
 1− x 

exp(m sin −1 x)
18. Find ∫x
2
dx on (–1, 1) where m is a real number. (Here for Y ∈ R, exp.(y)
1− x2
stands for ey).
Sol. Let t = sin–1 x, then
1
x = sin t, dt = dx , for x ∈ (–1, 1)
1 − x2

exp(m sin −1 x)
Hence ∫ x 2 dx = ∫ e mt sin 2 tdt
1− x 2

 1 − cos 2t 
= ∫ emt   dt
 2 

1 mt 1
=
2 ∫ e dt − ∫ emt ⋅ cos 2t dt + c
2
…(1)

Case (i): m = 0

exp(m sin −1 x)
From (1) ∫ x 2 dx
1− x 2

1 1
=
2 ∫ dt − ∫ cos 2t dt + C
2
t sin 2t
= − +C
2 4
sin −1 x 1
= − sin(2sin −1 x) + C
2 4

Case (ii):m ≠ 0

exp(m sin −1 x)
From (1) ∫ x 2 dx
1− x 2

1 e mt 1 e mt
= − (m cos 2t + 2 sin 2t) + C1
2 m 2 m2 + 4

e mt 1 1 
= m− 2 (m cos 2t + 2 sin 2t)  + C1
2  m +4 
−1
e m sin x
1 1 
= m− 2 (m cos(2sin −1 x) + 2sin(2 sin −1 x)  + C1
2  m +4 

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Integration of Some Special Types of Functions

px + q d
Type I: If the integral is of the form ∫ ax 2 + bx + c dx then take px + q = A (ax 2 + bx + c) + B .
dx

px + q d
Type II: ∫ dx . Take px + q = A (ax 2 + bx + c) + B .
ax 2 + bx + c dx

d
Type III: ∫ (px + q) ax 2 + bx + c dx . Take px + q = A (ax 2 + bx + c) + B .
dx

1 1
Type IV: ∫ dx . To evaluate this put px + q = .
(px + q) ax 2 + bx + c t

1 1
Type V: ∫ dx . To evaluate this, put x = .
(ax + b) cx + d
2 2 t

px + q ax + b 1
Type VI : ∫
ax + b
dx or ∫ px + q
dx or ∫ (px + q) ax + b dx or ∫
(px + q) ax + b
dx . Put ax + b = t2.

1
Then dx = 2t dt .
a

Integration Of Functions Which Are Rational in sin x and cos x.

dx dx
Type I:If the integral is of the form ∫ a 2 cos2 x + b2 sin 2 x or ∫ a cos2 x + b sin x cos x + c sin 2 x then

multiply both numerator and denominator with sec2 x and take tan x = t.
dx dx dx
Type II: If the integral is of the form ∫ a + b cos x or ∫ a + bsin x or ∫ a cos x + bsin x + c , take

x 1 x 2dt 2 tan x / 2 2t
tan = t ⇒ sec2 dx = dt ⇒ (1 + tan 2 x / 2)dx = 2dt ⇒ dx = . sin x = = ,
2 2 2 1+ t2 1 + tan x / 2 1 + t 2
2

1 − tan 2 x / 2 1 − t 2
cos x = = .
1 + tan 2 x / 2 1 + t 2

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a cos x + b sin x d
Type III :If the integral is of the form ∫ c cos x + d sin x dx , take a cos x + b sin x = A dx (c cos x + d
sin x). By equating the coefficients of cos x, sin x we get the values of A and B. Then the given
integral becomes A log|ccos x + d sin x| + Bx + k.

Very Short Answer Questions

Evaluate the following integrals.

dx
1. ∫
2x − 3x 2 + 1

dx
Sol. ∫
2x − 3x 2 + 1

dx 1 dx
=∫ = ∫
 2x 1 3 2
2  1
2
3 − x2 +  − x −
 3 3    
3  3

 1
x− 
1 −1  3 + C = 1 sin −1  3x − 1  + C
= sin    
3  2  3  2 
 3 

sin θ
2. ∫ dθ
2 − cos 2 θ

sin θ
Sol. ∫ dθ
2 − cos 2 θ

Put cos θ = t ⇒ − sin θ dθ = dt

dt dt
= ∫− = −∫
2 − t2 ( 2) 2 − t 2

 t  −1  cos θ 
= − sin −1   + C = − sin  +C
 2  2 

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cos x
3. ∫ sin 2 x + 4 sin x + 5 dx
cos x
Sol. ∫ dx
sin 2 x + 4 sin x + 5

put sin x = t ⇒ cos xdx = dt

dt dt
=∫ =∫
t 2 + 4t + 5 (t + 2) 2 + 1
= tan −1 (t + 2) + C = tan −1 (sin x + 2) + C

dx
4. ∫ 1 + cos2 x
dx sec 2 dx sec 2 xdx
Sol. ∫ =∫ =∫
1 + cos 2 x sec 2 x + 1 tan 2 x + 2

Let tan x = t ⇒ sec 2 x dx = dt

dt 1  t 
=∫ = tan −1  +C
t + ( 2)
2 2
2  2

1  tan x 
= tan −1  +C
2  2 

dx
5. ∫ 2 sin 2 x + 3cos2 x
dx sec 2 xdx
Sol. ∫ =∫
2 sin 2 x + 3cos 2 x 2 tan 2 x + 3

Let tan x = t ⇒ sec 2 x dx = dt

dt 1 dt
=∫
2∫
=
2t + 3
2
 3
2
t +
2

 2

2  2t 
= tan −1  +C
3  3 
2
2

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1  2 tan x 
= tan −1  +C
2 3  3 
1  2 
= tan −1  tan x  + C
6  3 

1
6. ∫ 1 + tan x dx
1
Sol. ∫ dx
1 + tan x

1 cos x dx 1 2 cos x dx
=∫ dx = ∫ = ∫
1+
sin x sin x + cos x 2 sin x + cos x
cos x

1 (cos x + sin x) + (cos x − sin x)

2∫
= dx
sin x + cos x

1 1 cos x − sin x
=
2 ∫ dx + ∫
2 sin x + cos x
dx

1 1
= x + log | sin x + cos x | +C
2 2

1
7. ∫ 1 − cot x dx
1 1 sin xdx
Sol. ∫ dx = ∫ dx = ∫
1 − cot x 1−
cos x sin x − cos x
sin x

1 (sin x − cos x) + (cos x + sin x)

2∫
= dx
sin x − cos x

1 1 cos x + sin x
=
2 ∫ dx + ∫
2 sin x − cos x
dx

1 1
= x + log | sin x − cos x | +C
2 2

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Short Answer Questions

Evaluate the following integrals.

1. ∫ 1 + 3x − x 2 dx

∫ 1 + 3x − x 2 dx = ∫ 1 − (x 2 − 3x)dx
Sol.
3 9
= ∫ 1 − (x − ) 2 − dx
2 2

2
 13   3
= ∫  −
  x − 
 2   2

 3  3
 x −  1 + 3x − x
2
 x− 
=
2 13
+ sin −1  2 +C
2 8  13 
 
 2 

(2x − 3) 1 + 3x − x 2 13 −1  2x − 3 
= + sin  +C
2 8  13 

9 cos x − sin x
2. ∫ 4sin x + 5cos x dx
9 cos x − sin x
Sol. ∫ dx
4sin x + 5cos x

d
let 9cosx-sinx =A ( 4sin x + 5cos x ) + B ( 4sin x + 5cos x )
dx
9cosx-sinx =A ( 4 cos x − 5sin x ) + B ( 4 sin x + 5 cos x )

Comparing the coefficients of sin and cos , we get

9 = 4A+5B and -5 = -5A+4B

Solving these equations , A =1 and B=1.

∴ 9cosx-sinx =1( 4 cos x − 5sin x ) + 1( 4 sin x + 5cos x )
=1( 4 cos x − 5sin x ) + 1( 4sin x + 5 cos x )

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9 cos x − sin x (4sin x + 5cos x) + (4 cos x − 5sin x)

∫ 4sin x + 5cos x dx = ∫ 4sin x + 5cos x
dx

4cos x − 5sin x
= ∫ dx + ∫ dx
4sin x + 5cos x
= x + log | 4sin x + 5cos x | +C

2cos x + 3sin x
3. ∫ 4 cos x + 5sin x dx
Sol. Let 2 cos x + 3 sin x = A(4 cosx + 5 sin x) + B(–4 sin x + 5 cos x)
Equating the coefficient of sinx and cosx, we get 4A + 5B = 2, 5A – 4B = 3.
A B 1
+5 −2 4 +5
−4 −3 5 −4

A B 1
= =
−15 − 8 −10 + 12 −16 − 25

23 2
A= ,B = −
41 41

2 cos x + 3sin x
∫ 4 cos x + 5sin x dx =
23 2 −4sin x + 5cos x
=
41 ∫ dx − ∫
41 4 cos x + 5sin x
dx

23 2
= x − log | 4 cos x + 5sin x | +C
41 41

dx
4. ∫ 1 + sin x + cos x
dx
Sol. ∫
1 + sin x + cos x

dx
= ∫ x x
 2 tan 1 − tan 2
2 2
1 + +
x
2 x
 1 + tan 1 + tan 2 
 2 2

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x
dx sec2
=∫ 2
x x x
1 + tan 2 + 2 tan + 1 − tan 2
2 2 2

x
sec 2 x 1 x
=∫ 2 put tan = t ⇒ sec2 dx = dt
x 2 2 2
2 + 2 tan
2

dt dt
= 2∫ =∫ log |1 + t | +C
2 + 2t 1+ t

x
= log 1 + tan +C
2

dx
5. ∫ 3x 2 + x + 1
dx dx
Sol. ∫ =∫
3x + x + 1
2  1 1
3 x2 + x + 
 3 3

1 dx 1 dx
= ∫
3  2
1 1 1 3 
= ∫
2 2
+ + − 1   11 


x 
6  3 36 x +  + 
 6  6 

1 1  x + (1/ 6) 
= ⋅ tan −1  +C
3 11  ( 11/ 6) 
 
6

2  6x + 1 
= tan −1  +C
11  11 

dx
6. ∫
5 − 2x 2 + 4x

dx
Sol. ∫
5 − 2x 2 + 4x

5 – 2x2 + 4x

 5  5
= −2  x 2 − 2x −  = −2 (x − 1) 2 − 1 − 
 2  2

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 7 7 
= −2 (x − 1) 2 −  = 2  − (x − 1)2 
 2 2 

1
Now ∫ dx
5 − 2x 2 + 4x

1
=∫ dx
7 
2  − (x − 1) 2 
2 
1 1
=
2 ∫ 2
dx
7
  − (x − 1)
2
2

1 (x − 1)
= sin −1 +C
2 7/2
1 2
= sin −1 (x − 1) + C
2 7

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Long Answer Questions

Evaluate the following integrals.

x +1
1. ∫ dx
x − x +1
2

Sol. take x+1 =A

d 2
dx
( )
x − x +1 + B

x+1 =A ( 2x − 1) + B

Comparing the coefficients of like terms,

2A = 1 and B-A =1
1 3
⇒ A= and b =
2 2

1 3
∴ x+1 = ( 2x − 1) +
2 2

1 3
(2x − 1) +
x +1
∫ dx = ∫ 2 2 dx
x − x +1
2
x − x +1
2

1 (2x − 1)dx 3 dx
=
2 ∫ + ∫
x − x +1 2
2
x − x +1
2

3 dx
2∫
= x2 − x +1 +
2
1  3
2

 x −  + 
 2  2 

 1
3 x− 
= x 2 − x + 1 + sinh −1  2 +C
2  3 
 
 2 

3  2x − 1 
= x 2 − x + 1 + sinh −1  +C
2  3 

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2. ∫ (6x + 5) 6 − 2x 2 + x dx

Sol.

let 6x+5=A
d
dx
( )
6 − 2x 2 + x + B

⇒ 6x + 5 = A(1 – 4x) + B

Equating the coefficients

−3
6 = −4A ⇒ A =
2

Equating the constants

A+B=5
3 13
B = 5 – A = 5+ =
2 2

∫ (6x + 5) 6 − 2x 2 + x dx

3 13
=−
2 ∫ (1 − 4x) 6 − 2x 2 + x dx +
2
6 − 2x 2 + x dx

3 (6 − 2x 2 + x)3/ 2 13 x
=− + 2 ∫ 3 − x 2 + dx
2 3/ 2 2 2

2 2
13 7  1
= −(6 − 2x 2 + x)3/ 2 +
2 ∫   −  x −  dx
4  4

13
= −(6 − 2x 2 + x)3/ 2 +
2
 1 2 x  1
 x −  3− x +  x− 
 4 2 49 −1 
+ sin
4
+C
 2 32 7 
   
 4 

13  (4x − 1) 6 − 2x 2 + x 49 −1  4x − 1  
= −(6 − 2x 2 + x)3/2 +  + sin   + C
2 16 × 2 32  7 
 

13 637  4x − 1 
= −(6 − 2x 2 + x)3/2 + (4x − 1) 6 − 2x 2 + x + sin −1  +C
16 32 2  7 

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dx
3. ∫ 4 + 5sin x
dx dx
Sol. ∫
4 + 5sin x ∫
=
x
2 tan
4+5 2
x
1 + tan 2
2

x x 1
put tan = t ⇒ sec2 ⋅ dx = dt
2 2 2

2dt 2dt
⇒ dx = =
sec2
x 1+ t2
2

dt dt
G.I. = 2 ∫ = 2∫
1+ t 2
4 + 4t 2 + 10t
2t
4+5
1+ t2

1 dt 1 dt
= ∫ = ∫
2 t 2 + 5t + 1 2  5  2  3  2
2 t +  − 
 4 4

5 3
t+ −
1 1 4 4 +C
= log
2 2⋅ 3 5 3
t+ +
4 4 4

1 4t + 2 1 2t + 1
= log + C = log +C
3 4t + 8 3 2t + 4

x
+1
2 tan
1 2
= log +C
3  x
2  tan  + 2
 2

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1
4. ∫ 2 − 3cos 2x dx
1 dx
Sol. ∫ dx = ∫
2 − 3cos 2x 1 − tan 2 x
2−3
1 + tan 2 x

put tan x = t ⇒ sec 2 x dx = dt

dt
dx =
1+ t2

dt dt
GI = ∫ =∫
1+ t 2
2 + 2t − 3 + 3t 2
2

1− t2
2−3
1 + t2

dt 1 dt
=∫
5∫
=
5t − 1
2
 1 
2
t2 −  
 5

1
t−
1 (1/ 2) 5 +C
= log
5 5 1
t+
5

1 5t − 1
= log +C
2 5 5t + 1

1 5 tan x − 1
= log +C
2 5 5 tan x + 1

5. ∫x 1 + x − x 2 dx

Sol. Let x = A(1 – 2x) + B

Equating the coefficients of x

1 = –2A ⇒ A = –1/2
Equating the constants

0 = A + B ⇒B = –A = 1/2

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

2 2
1 (1 + x − x 2 )3/ 2 1  5  1
=− + ∫   −  x −  dx
2 3/ 2 2  2   2

 1  1 
  x −  1+ x − x2  x − 
= − (1 + x − x 2 )3/2 +  
1 1 2 25
+ sin −1  2 
3 2 2 8  5 
  
  2 

1 (2x − 1) 1 + x − x 2 5  2x − 1 
= − (1 + x − x 2 )3/2 + + sin −1  +C
3 8 16  5 

dx
6. ∫
(1 + x) 3 + 2x − x 2

dx dx
Sol. ∫ =∫
(1 + x) 3 + 2x − x 2 (1 + x) (3 − x)(1 + x)

Put 1 + x = t 2 ⇒ dx = 2t dt

2t dt 2dt 2 dt
G.I. = ∫ =∫ =∫
t 2 t 2 (4 − t 2 ) t2 4 − t2 t3 4
−1
t2

4 8
Put 2
− 1 = y2 ⇒ − dt = 2y dy
t t3

2 y
⇒ 3
dt = − dy
t 4

y dy 1 1
G.I. = 2 ∫ − = − ∫ dy = − y + C
4 y 2 2 2

1 4
=− −1 + C
2 t2
1 4 1 3− x
=− −1 + C − +C
2 1+ x 2 3+ x

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dx
7. ∫ 4 cos x + 3sin x
dx dx
Sol. ∫
4 cos x + 3sin x ∫ 1 − tan 2
=
x x
2 tan
4 2 +3 2
x 2 x
1 + tan 2 1 + tan
2 2

x 2dt
Put tan = t ⇒ dx =
2 1+ t2

2dt
I=∫ 1+ t2 = 2∫
dt
(1 − t ) 3 ⋅ 2t
2
4 − 4t 2 + 6t
4 +
1+ t2 1+ t2

1 dt 1 dt
=− ∫
2 t2 − 3 t −1
=− ∫
2  3 2  5 2
2 t −  − 
 4 4

3 5
t− −
1 1 4 4 +C
=− log
2 2⋅ 5 3 5
t− +
4 4 4

1 t−2 1 2t − 4
= − log + C = − log +C
5 t + (1/ 2) 5 2t + 1

 x 
2  tan − 2 
= − log   +C
1 2
5 x
2 tan + 1
2

1
8. ∫ sin x + 3 cos x
dx

x 2dt
Sol. Let t = tan so that dx =
2 1+ t2

2t 1 − t2
sin x = , cos x =
1+ t 2
1+ t2

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dt
2
I=∫ 1+ t2 = 2∫
dt
2t 3(1 − t )
2
3(1 − t 2 ) + 2t
+
1 + t2 1+ t2

2 dt 2 dt
= ∫
3 1− t2 + 2 t
= ∫ 2
3  2   1 
2

3   −t − 
 3  3

2 1
+t−
2 1/ 4 3 3 +C
= log
3 3 2 1
−t+
3 3

1
t+
1 3 + C = 1 log 3t + 1
= log +C
2 3−t 2 3( 3 − t)

x
3 tan +1
1 2
= log +c
2  x
3  3 − tan 
 2

dx
9. ∫ 5 + 4 cos 2x
dt
Sol. t = tan x ⇒ dt = sec2 x dx, dx =
1+ t2

dt
I = ∫ 1+ t 2 = ∫
2 dt
5 + 4(1 − t ) 5 + 5t + 4 − 4t 2
2

1+ t2

dt 1 t
=∫ = tan −1   + C
t +9 3
2
 3

1  tan x 
= tan −1  +C
3  3 

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2sin x + 3cos x + 4
10. ∫ 3sin x + 4 cos x + 5 dx
Sol.
2sin x + 3cos x + 4
∫ 3sin x + 4 cos x + 5 dx
Let 2 sinx + 3 cos x + 4 = A ( 3sinx + 4cosx + 5 ) + B
d
( 3sinx + 4cosx + 5) + C
dx

2 sinx + 3 cos x + 4
= A(3sinx + 4cosx + 5) + 3(3cosx – 4sinx) + C
Equating the coefficients of
sin x and cos x,
we get 3A – 4B = 2
and 4A + 3B = 3
Solving these equations,
18 1
A= , B=
25 25

Equating the constants

4 = 5A + C
18 2
C = 4 – 5A = 4 – 5 ⋅ =
25 5
18 1 2
2 sinx + 3 cos x + 4 = ( 3sinx + 4cosx + 5) + ( 3cosx − 4sinx ) +
25 25 5

2sin x + 3cos x + 4
∴∫ dx
3sin x + 4cos x + 5

18 1 3cos x − 4sin x
=
25 ∫ dx + ∫
25 3sin x + 4 cos x + 5

2 dx
+ ∫
5 3sin x + 4 cos x + 5

18 1
= x + log | 3sin x + 4 cos x + 5 |
25 25
2 dx
+ ∫
5 3sin x + 4 cos x + 5
...(1)

dx
Let I = ∫
3sin x + 4 cos x + 5

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2dt
x 2dt 1 + t2
Put tan = t ⇒ dx =
2 1+ t2
I = ∫ 3 − 2t 4(1 + t 2 )
+ +5
1+ t2 1+ t2

dt dt
= 2∫ = 2∫
6t + 4 − 4t + 5 + 5t
2 2
t + 6t + 9
2

dt 2 2
= 2∫ =− =−
(t + 3) 2 t +3 3 + tan
x
2

Substituting in (1)
18 1
I= ⋅ x + log | 3sin x + 4 cos x + 5 |
25 25
4
− +C
 x
5  3 + tan 
 2

5−x
11. ∫ dx on (2, 5).
x−2

5−x (5 − x)2
Sol: ∫ dx = ∫ dx
x−2 (x − 2) (5 − x)

5−x
=∫ dx
(x − 2) (5 − x)

5−x
=∫ dx
5x − x 2 − 10 + 2x

5−x
=∫ dx
7x − x 2 − 10

d
Let 5 − x = A (7x − x 2 − 10) + B
dx

= A(7 − 2x) + B

Equating coefficient of x and constant terms on both sides we get

7 3
–2A = –1 ⇒ A = 1/2 and 7A + B = 5 ⇒ B = 5 – 7A = 5 − = .
2 2

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1 3
∴5 − x = (7 − 2x) +
2 2
5−x
∴∫ dx
7x − x 2 − 10

1 (7 − 2x) 3 dx
= ∫ dx + ∫ …(1)
2 7x − x 2 − 10 2 7x − x 2 − 10

Consider the first integral on RHS and suppose 7x – x2 – 10 = t ⇒ (7 – 2x)d = dt

1 (7 − 2x) 1 dt 1 −1/ 2
∴ ∫ dx = ∫ = ∫ t dt
2 7x − x − 10
2 2 t 2

1 t1/ 2
= = t = 7x − x 2 − 10 …(2)
2 1/ 2

Consider the second integral and

7x − x 2 − 10 = −(x 2 − 7x + 10)
 7 49 49 
= −  x2 − 2   x + − + 10 
 2 4 4 
 7  9
2
= −  x −  − 
 2  4 

 7 3 
2 2
= −  x −  −   
 2   2  
2 2
3  7
=   −x − 
2  2

dx dx
∴∫ =∫
7x − x 2 − 10 2
3  7
2

  −x − 
2  2

 7
 x− 
= sin −1  2 = sin −1  2x − 7  …(3)
3  
 3 

 
 2 

∴ From (1), (2) and (3)

5−x
∫ dx =
7x − x 2 − 10
3  2x − 7 
7x − x 2 − 10 + sin −1  +c
2  3 

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1+ x
12. ∫ dx on ( −1,1)
1− x

1+ x (1 + x) 2
Sol: ∫ dx = ∫ dx
1− x 1− x

1+ x dx xdx
=∫ dx = ∫ +∫
1− x 2
1− x2 1− x2
1 ( −2x)dx
= sin −1 x −
2 ∫ 1− x2

= sin −1 x − 1 − x 2 + c

dx
13. ∫ on (–1, 3).
(1 − x) 3 − 2x − x 2

1
Sol: Put 1 – x = 1/t ⇒ x = 1 −
t

1
∴ dx = + dt
t2
2
 1  1
Also 3 − 2x − x 2 = 3 − 1 −  − 1 − 
t   
t 

2 1 2 
= 3+ − 2 −  2 − + 1
t t t 
4 1 4t − 1
= − = 2
t t2 t

1
 2  dt
=∫  
dx t
∴∫
(1 − x) 3 − 2x − x 2  1  4t − 1
 
t t2

dt 2 4t − 1
=∫ = +c
4t − 1 4
1
= 4t − 1 + c
2

1  1 
= 4  −1 + c
2  1− x 

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1 4 −1+ x 1 3+ x
= +c = .
2 1− x 2 1− x

dx
14. ∫ ( x + 2) x +1

dx
Sol. ∫ ( x + 2) x +1

put x+1 =t 2

dx = 2tdt and x+2=1+t 2

2t
G .I . = ∫ dt
(1 + t 2 ) t
= 2. tan −1 t + c = 2 tan −1 x + 1 + c

dx
15. ∫ ( 2 x + 3) x+2

Sol. put x+2 =t 2

dx = 2tdt and 2x+3=2t 2 − 1

2t 1
G.I . = ∫ dt = 2 ∫ dt
( 2t − 1) t
2
( 2t )
2
−1

1 2 t −1
= log +c
2 2 t +1

1 2x + 4 −1
= log +c
2 2x + 4 + 1

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dx
16. ∫
(1 + x ) x − x2

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

put x=sin 2t ⇒ dx = 2sin t.cos t.dt

1
dt = 2. ( tan t − sec t ) + c
2 sin t.cos t.dt
=∫ = 2∫
(1 + sin t ) sin t.cos t 1 + sin t

1
= 2∫ dt = 2. ( tan t − sec t ) + c
1 + sin t

dx
17. ∫ ( x + 1) 2 x 2 + 3x + 1

dx
Sol. ∫ ( x + 1) 2 x 2 + 3x + 1

1 1− t −1
put x+1 = ⇒ x = and dx= 2 dt
t t t

1 −1
g.i. = ∫ . dt
1 1− t   1− t 
2 t2
. 2  + 3  +1
t  t   t 

−1
=∫ dt
2 + 2t 2 − 4t + 3t − 3t 2 + t 2

1
= −∫ dt = 2 2 − t + c
2−t

1 2x +1
= 2 2− +c = 2 +c
x +1 x +1

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18. ∫ e x − 4dx

ex − 4
Sol. ∫ e x − 4dx = ∫
ex − 4
dx

ex 4
=∫ dx − ∫ dx
ex − 4 ex − 4

e− x/ 2
= 2 e − 4 − 4∫
x
dx
1 − 4e − x

e− x /2
= 2 e x − 4 + 4∫ dx
1− 4 (e )
− x/ 2 2

e− x /2
= 2 e − 4 + 4.∫
x
dx
1 − ( 2e − x /2 )
2

= 2 e x − 4 + 4sin −1 2e− x / 2 + c
−1 2e− x /2
= 2 e − 4 + 4 tan
x
+c
1 − 4e − x

2
= 2 e x − 4 + 4 tan −1 +c
e −4
x

19. ∫ 1 + sec xdx

x
2 cos 2
1 + cos x 2 dx
∫ 1 + sec xdx = ∫
cos x
dx = ∫
x
1 − 2 sin 2
2

x 1 x
2.cos 2.cos
=∫ 2 dx = 2 ∫ 2 2 dx
2 2
 x  x
1 −  2 sin  1 −  2 sin 
 2  2

 x
= 2sin −1  2 sin  + c
 2
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1
20. ∫ 1 + x4
dx

1
Sol. ∫ 1 + x4
dx

1 2 1 1 + x2 + 1 − x2
2 ∫ 1 + x4 2∫
= dx = dx
1 + x4

1  1 + x2 1 − x2 
= ∫  + 4 
dx
1+ x 1+ x 
4
2

 1 1 
1+ 2 −1
1  x + x2 
= ∫ dx
2  x2 + 1 x2 + 1 
 x2 x2 

   1 1 
1 1 x− x+ − 2 
 1+ 2 1−  1 1 x− 1 x
= tan −1 +c
dx  2  2
1 x x2 log
2∫ 
= dx − ∫ 2 2 2 1
  x+ + 2 
( ) ( 2)
2 2
1 2
 1 2
 
 x−  + 2 x+  −  x
  x  x 

1  −1 x 2 − 1 1 x2 + 1 − 2 
=  tan − log 2 +c
2 2 x 2 2 x + 1 + 2 

dx
21. Evaluate ∫ 5 + 4 cos x .

x 2 dt
Sol: Put tan = t then dx =
2 1+ t2

1− t2
And cos x =
1+ t2

2dt
dx 1+ t2
∫ 5 + 4 cos x = ∫  1− t2 
5 + 4 2 
 1+ t 

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2dt dt
=∫ = 2∫ 2 2
9+ t 2
3 +t
1 t
= 2   tan −1  
3  3
 x
 tan 
2
= tan −1  2 +c

3  3 

dx
22. ∫ 3cos x + 4sin x + 6
x 2 dt
Sol: Let tan = t then dx =
2 1+ t2

2t 1− t2
sin x = and cos x =
1+ t2 1 + t2

dx
∴∫
3cos x + 4sin x + 6

2dt
=∫ 1+ t2
 1 − t 2   2t 
3 2 
+ 4 2 
+6
 1+ t   1+ t 

2 dt
=∫
3 − 3t + 8t + 6 + 6t 2
2

2dt
=∫ 2
3t + 8t + 9
2 dt
= ∫
3 t2 + 8 t + 3
3
2 dt
= ∫
3  4 2 16
t + 3  +3− 9
 
2 dt
= ∫
3  4 2 11
t+  +
 3 9

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2 dt
= ∫
3  4  2  11  2
t +  + 
 3  3 
 4
2 3 t+ 
= tan −1  3
3 11  11 
 
 3 

2  3t + 4 
= Tan −1   + c.
11  11 

1
23. Evaluate ∫ dx on I = (0, 1).
sin −1 x 1 − x 2

1
Sol: Let sin–1 x = t then dx = dt
1− x2

1 1
∴∫ dx = ∫ dt
−1
sin x 1 − x 2 t

= ∫ t −1/ 2 dt

= 2 t + c = 2 sin −1 x + c .

x
24. Find ∫ dx , x ∈ I = (0, 1).
1− x

Sol: Let 1 – x = t2 over (0, 1)

Then –dx = 2t dt and x = 1 – t2

x (1 − t 2 )2tdt
∴∫ dx = − ∫
1− x t

 t3 
= −2 ∫ (1 − t) 2 dt = −2  t − 
 3
 (1 − x)3/ 2 
= −2  1 − x − 
 3 
2
= (1 − x)3 / 2 − 2 1 − x + c
3

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dx
25. Evaluate ∫ (x + 5) on (−4, ∞) .
x+4

Sol: Let x + 4 = t2 then dx = 2t dt

Defined over (–4, ∞)

dx 2t dt dt
∴∫ =∫ 2 = 2∫ 2
(x + 5) x + 4 (t + 1)t t +1
= 2 tan −1 t + c
= 2 tan −1 ( x + 4) + c

dx
26. Evaluate ∫ .
x 2 + 2x + 10

Sol: x 2 + 2x + 10 = x 2 + 2x + 1 + 9

= (x + 1) 2 + 32

dx dx
∴∫ =∫
x 2 + 2x + 10 (x + 1) 2 + 32

Take x + 1 = t ⇒ dx = dt

dt t
=∫ = sinh −1   + c
t 2 + 32 3

 x +1 
= sinh −1  +c
 3 

 dx  x 
 use ∫ = sinh −1    .
 a 2 − x2  a  

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