1

Lakshya JEE 2.0 (2024)

Indefinite Integration DPP-01

1 − tan 2 x x4 + x2 + 1
1.  1 + tan 2 x dx 8.  x2 − x + 1
dx is equal to

x3 x 2
dx 1) − + x+C
2.  1 + cos x 3 2
x3 x 2
(2) + + x+C
3 2
dx
3.  sin 2 x cos 2 x (3)
x3 x 2
− − x+C
3 2

x3 x3 x 2
(4) + − x+C
4.  x + 1 dx 3 2

2  e x dx =
x
x4 + x2 + 1 9.

( )
5. dx
2 1 + x2 (2e) x e2
(1) +c (2) +c
ln ( 2e ) ln ( 2e )

x 4 + 3x 2 + 1 (2e) x
6.  x
dx equals (3)
ln2 x
+c (4) None of these

7 3 1
2 2 −
(1) x + 2x 2 + x 2 + C 2 x +1 − 5 x −1
2  ax 1 bx
7
7 3
10.  10 x dx =
−
ln a 5 ln b
+ c , where c is an
2 2 2 2
(2) x + x + x +C integration constant then 10ab =
9 3 (1) –1 (2) 0
5 3 1
2 2 3 2 (3) 1 (4) 2
(3) x + x + 2x 2 + C
5 2
9 5 1 ( x 2 + sin 2 x)sec2 x
(4)
2 2 6 2
x + x + 2x 2 + C 11.  1 + x2
dx =
9 5
(1) tan x + tan −1 x + c
2 + 3x 2 (2) tan −1 x − tan x + c
7.  x2 (1 + x2 ) dx =
(3) tan x − tan −1 x + c
(1) tan −1 x +
2
+c (4) c − tanx − tan −1x
x
2 sin2 x + sin5 x − sin3 x
− tan −1 x + c
(2)
x 12.  cosx + 1 − 2sin 2 2 x
dx =
1
(3) tan −1 x − + c (1) 2cosx + c
x (2) −2sinx + c
2 (3) 2sinx + c
(4) tan −1 x − + c
x (4) −2cosx + c
 2

x6 − 1 ( x + 1) ( x 2 – x )
13.  x2 + 1 dx = 15.  x x +x+ x
dx

x5 x3
(1) − − + x − 2 tan −1 x + c
cos 4 x + 1
 cot x – tan x dx
5 3 16.
x5 x3
(2) + + x − 2tan −1 x + c
5 3
17. A function g defined for all positive real numbers,
x5 x3
(3) − + x − 2tan −1 x + c satisfies g(x2) = x3 for all x> 0 and g (1) = 1.
5 3
Compute g (4).
x5 x3
(4) − + x + 2tan −1 x + c
5 3
18.  sin x cos x cos2x cos4x dx
e2 x – 1
14.  ex
dx
 1

Lakshya JEE 2.0(2024)

Indefinite Integration DPP-02

2x – 1 sin( nx)
1.  x–2
dx 11.  x
dx

sin 2 x + sin 5 x – sin 3 x x6 – 1

2.  cos x + 1 – 2sin 2 2 x
dx 12.  x 2 + 1 dx

cos4 x – sin 4 x sin 3 x + cos3 x

3.  1 + cos 4 x dx(cos 2 x  0) 13.  dx
sin 2 x cos 2 x

4.  cos x dx x4 + x2 + 1
14.  2(1 + x 2 )
dx

x 21
5.  4cos 2 .cos x.sin x dx

2 15. 1– sin 2 x dx

cos 2 x
6.  cos2 x sin 2 x dx –1  1 + cos x 
16.  tan   dx
 sin x 

 (3sin x cos
2
7. x – sin 3 x) dx
 2  9 x  7 x  
17.  sin  +  – sin 2 
 8 4
+   dx
 8 4 
cos x – sin x
8.  cos x + sin x (2 + 2sin 2 x) dx
cos8 x – cos7 x
18.  1 + 2cos5 x
dx
(1 + x) 2
9.  x(1 + x 2 ) dx
 (tan
3
19. x – x tan 2 x) dx

2sin( x 2 + 1) – sin 2( x 2 + 1)
10.
sec2 x
 1 + tan x dx
20.  x. 2sin( x 2 + 1) + sin 2( x 2 + 1)
dx
 1

Lakshya JEE 2.0(2024)

Indefinite Integration DPP-03

dx dx
1.  7 − x2
= 4.  4 x2 + 9
 x  1 −1  2 x 
(1) − sin −1  +c (1) tan   + c
 7 4  3
 x  1 −1  2 x 
(2) cos−1  +c (2) tan   + c
 7 8  3 
 x  1  2x 
(3) − cos−1  +c (3) tan −1   + c
 7 12  3 
 x  1 −1  2 x 
(4) sin −1  +c (4) tan   + c
 7 6  3

e x dx dx 1
2.  5.  cosx − sinx = log f ( x ) + c , then f ( x )
e2 x − 1 2

(1) (
ln ex + e2 x − 1 + c ) equals:

(1) tan
x
(2) cot
x

(2) (
–ln ex + e2 x − 1 + c ) 2
 x 3 
(3) tan  + 
2
 x 3 
(4) tan  + 
(3) (
ln ex − e2 x − 1 + c ) 2 8  2 5 

(4) ln ( e + x
e2 x + 1) + c 6. If f(0) = f' (0) = 0 and f" (x) = tan2x then f(x) is
1 1
(1) log sec x − x2 (2) logcos x + x2
2 2
e x dx
 4 + e2 x
1
3. (3) log sec x + x2 (4) x4 + x3 + 1
2
1 −1  e x 
(1) − tan   + c
 2 x2 tan −1 x3
 1 + x6 dx =
2   7.
1 −1  e x 
tan   + c
( )
1 2
(2)
2  4 (1) tan −1 x3 + c
  6
1 −1  e x  ( )
1 2
(3) tan   + c (2) − tan −1 x3 + c
2  2 6
 
( )
1 2
tan −1 x3 + c
1 −1  e x  (3)
4
(4) tan   + c
6
( )
2   1 2
(4) tan −1 x3 + c
3
 2

dx cos ec(tan −1 x)
8.  1 − 4 x2
= k1 f ( x) + k2 where 𝑘1 is constant and 12.  1 + x2 dx
k 2 is an integration constant then 2k1 = (1) ln(1 − x) + c
(1) 1 (2) 2  1
(2) ln 1 + x −  + c
(3) 3 (4) 4  x
 1 + x2 − 1 
9.  cosec xdx =
4
(3) ln  +c
 x 
 
cot 3 x
(1) cot x − +c (4) ln x + c
3
cot 3 x
(2) − cot x − +c e x dx
3 13.  5 − 4e x + e2 x
3

( )
cot x
(3) cot x + +c
3 (1) ln e x − 2 + 5 − 4e x + e2 x + c
3

( )
cot x
(4) − cot x + +c
3 (2) − ln e x − 2 + 5 − 4e x + e2 x + c

10.  tan
4
x dx = (
(3) ln e x + 2 + 5 − 4e x + e2 x + c )
ln ( e − 2 − )+c
tan3 x
(1) − − tan x + x + c (4) x
5 − 4ex + e2 x
3
tan3 x
(2) − tan x − x + c
3 dx
tan3 x
14.  1 + ex
(3) − tan x + x + c
3
(1) − log | e(−x) + 1| +c
3
tan x
(4) + tan x + x + c (2) log | e(− x) + 1| +c
3
(3) − log | e(−x) −1| +c
sin 2 x dx
11.  9 − sin 4 x
= (4) log | e( x) + 1| +c

 sin 2 x 
(1) sin −1  +c
 3 
 
15. 10sin 2x  cos 2x cos2 2 x − 5dx
5
−1  sin x2 (1) (cos2 2 x − 5)3/2 + c
(2) − sin  +c 3
 3 

5
 sin 2 x  (2) (cos2 2 x + 5)3/2 + c
(3) sin −1  +c 3
 6 
  6
(3) − (cos2 2 x − 5)3/2 + c
 sin x  2 3
(4) sin −1  +c
 4 
  5
(4) − (cos2 2 x − 5)3/2 + c
3
 3

(2 x + 1) 4x + 3
16.  ( x2 + 4 x + 1)3/2 dx 19.  3x2 + 3x + 1 dx
x3 1 2
(1) +C (1) ln(3x2 + 3x + 1) + tan −1(2 3x + 3) + C
( x 2 + 4 x + 1)1/2 3 3
2 2
(2)
x
+C (2) ln(3x2 + 3x + 1) + tan −1(2 3x + 3) + C
3 3
( x + 4 x + 1)1/2
2
2 1
x2 (3) ln(3x2 + 3x + 1) + tan −1(2 3x + 3) + C
(3) +C 3 3
( x 2 + 4 x + 1)1/2
2 2
1 (4) ln(3x2 + 3x + 1) − tan−1(2 3x + 3) + C
(4) +C 3 3
( x + 4 x + 1)1/2
2

dx
(sin x )
−1 3 20.  2 + sin2 x
17.  1 − x2
dx
1  3 tan x 
(1) tan −1  +c
6  2 
xdx  3 tan x 
18.  x4 + x2 + 1 (2)
1
tan −1  +c
3  2 
2  1
(1) tan −1  x2 +  + C 1  3 tan x 
3  2 (3) − tan −1  +c
6  2 
1  2 x2 − 1 
(2) tan −1  +C  3 tan x 
 3  (4) −
1
tan −1 
3  +c
3  2 
1  2 x2 + 1 
(3) tan −1  +C
 3 
3 
1  1
(4) − tan −1  x2 +  + C
3  2
 1

Lakshya JEE 2.0 (2024)

Indefinite Integration DPP-04

1. 
dx
= 4.  cot  d 
x − 2x + 3
2

log x − 1 − x2 − 2x + 3 + c 2e x + 3e− x 
( )  + c
(1) 1 −x
 4e x + 7e− x dx = 14 ux + v ln | 4e + 7e |
x
5. If
(2) − log x − 1 + x2 − 2x + 3 + c
Then u + v is equal to:
(3) log x − 1 + x2 − 2x + 3 + c
x2 + 1
(4) log x + 1 + x2 − 2x + 3 + c
6.  x4 − x2 + 1 dx is equal to
 x2 − 1 
tan −1  +c
(1)
 x 
5x + 4  
2.  x + 2x + 5
2
dx
 x2 + 1 
tan −1  +c
(2)
 2 x 
 
(1) –5. x2 + 2x + 5
 x2 − 1 
− tan −1  +c
 x 
(3)
− ln x + 1 + x + 2x + 5 + c
2
 

5. x2 + 2 x + 5  x2 − 1 
(2) tan −1  +c
(4)
 2 x 
 
− ln x + 1 − x2 + 2x + 5 + c

(3) 5. x2 + 2 x + 5 x2 + 4 1 −1  x2 − 4 
7.  x4 + 16 k tan  kx  + c
dx =
− ln x + 1 + x2 + 2x + 5 + c  
(1) 2 (2) 4 2
(4) 5. x + 2 x + 5
2
(3) 2 2 (4) 2
+ ln x + 1 + x + 2x + 5 + c
2

e3x + e x
8.  e4x − e2x + 1 dx =
cos  d 
3. If  5 + 7sin  − 2cos2  d = Alog B() + c , then (1)
1
log(e4 x − e2 x + 1) + c
4
B()
can be: (2) tan−1(ex − e− x ) + c
A
2sin  + 1 2sin  + 1 (3) tan−1(ex + e− x ) + c
(1) (2)
5(sin  + 3) sin  + 3 (4) tan−1(e− x − ex ) + c
5(2sin  + 1) 5sin 
(3) (4)
sin  + 3 2sin  + 1
 2

9. For real numbers , ,  and  is 10. Suppose

 x2 + 1  A= 
dx
and B =  2
dx
( x2 − 1) + tan −1  
 x  dx
x + 6x + 25
2
x − 6x − 27
  x2 + 1   x +3 x −9
( x4 + 3x2 + 1) tan −1   If 12( A + B) =   tan −1   +   ln +C ,
 x   4  x+3
  x2 + 1   then the value of ( + ) is:
=   ln  tan −1 
  x  
   (1) 3
(2) 4
 ( x2 − 1)  −1  x + 1 
2
+ tan   +   tan   + c ,
 x (3) 5
   x 
(4) 6
where c is an arbitrary constant, then the value of
10( +  + ) =
 1

Lakshya JEE 2.0 (2024)

Indefinite Integration DPP-05

1. x log x dx is equal to: log x

5. dx
x2
x2 1
(1) (2log x 1) c (1) (1+logx) + C
4 2
x2 (2) x(1 + logx) + C
(2) (2log x 1) c (3) –x(1 + logx) + C
2
1
x2 (4) (1+logx) + C
(3) (2log x 1) c x
4
(4) None of these sin 1 xdx
6.
(1 x 2 )3/2
2
2. 2 x3 e x dx x sin 1
x
(1) log 1 x2 c
2 2 2
(1) x2 e x ex c 1 x
1
2 2 x sin x
(2) x2 e x ex c (2) log 1 x2 c
2 2 1 x2
(3) x2 e x ex c 1
x sin x
2 x 2
x 2 (3) log 1 x2 c
(4) x e e c 1 x 2

1
x sin x
3. 4 x cos(2 3x)dx (4) log 1 x2 c
2
1 x
4x 4
(1) sin(2 3x) cos(2 3x) c
3 9 7. x2x dx
4x 4
(2) sin(2 3x) cos(2 3x) c
3 9
8. The value of is ex ( x5 5x4 1)dx
4x 4
(3) sin(2 3x) cos(2 3x) c
3 9 (1) ex . x5 + C (2) ex . x5 + ex + C
4x 4
(4) sin(2 3x) cos(2 3x) c (3) ex + 1 .x5 + C (4) 5x4 . ex + C
3 9

4.
2
log x dx equals 9. e x (tan x logsec x)dx

(1) x(log x)2 – 2xlogx – 2x + C

(2) x(log x)2 – 2xlogx – x + C 2
(3) x(log x)2 – 2xlogx + 2x + C 10. Evaluate : xe x (sin x2 cos x2 )dx
(4) x(log x)2 – 2xlogx + x + C
11. sin(ln x) cos(ln x) dx
 2

ex 18. If x 2e 2x
dx e 2x
(ax2 + bx + c) + d, then the
12. 1 x ln x dx
x value of |a/bc| is
(1) 1 (2) 2
2 x
x e (3) 3 (4) 4
13. 2
dx
x 2
1 sin x x /2
19. e dx, x 0,
2
1 cos x 2
1 x
x
14. The value of e dx is x /2 x
1 x2 (1) e sin C
2
1 x 1 x x /2 x
(1) e x C (2) ex C (2) e sec
C
2
1 x2 1 x2
x
(3) e x/2 sin C
ex 2
(3) C (4) e (1 – x) + C
x
1 x2 (4) e x /2 cos
x
C
2

1 sin x
15. If ex dx f(x) + constant, then f(x) is 20. e3x sin 2 xdx equals
1 cos x
equal to e3 x
(1) (3 cos2x + 2sin2x) + C
x x 13
(1) e x cot C (2) e x
cot C
2 2 e3 x
(2) (3 sin2x – 2cos2x) + C
13
x x
(3) e x cot C (4) e x
cot C e3 x
2 2 (3) (–3sin2x – 2cos2x) + C
13
e3 x
x
1 (4) (3sin2x + 2cos2x) + C
1 x 13
16. 1 x e dx is equal to
x
1 1
x x
(1) ( x 1)e x C (2) xe x C
1 1
x x
(3) ( x 1)e x C (4) xe x C

2
x x 2 f ( x)e x
17. If e dx C the the sum f(x)
x 4 x 4
1
+ f2(x) …..  at, x is
2
1 1
(1) (2)
2 4
3
(3) 1 (4)
2
 1

Lakshya JEE 2.0 (2024)

Indefinite Integration DPP-06

1.  sin log x + coslog x dx is equal to

6. If 
2x + 3
dx = − a

 log
 ( x –1) 2 
 2 −
(1) sin(log x) + cos(log x) + C ( x –1)( x 2 + 1)   x +1 
1
(2) x sin(log x) + C tan −1 x + C , x  1 where C is any arbitrary
2
(3) x cos(log x) + C
constant, then the value of ‘a’ is
(4) x tan x + C 5 5
(1) (2) −
4 3
2.  2 x 2 − x + 1 dx (3) −
5
6
(4) −
5
4

f ( x)
3. Evaluate  (2 x − 4) 4 + 3 x − x 2 dx 7. If  x3 − 1 dx , where f(x) is a polynomial of degree
2 in x such that f(0) = f(1) = 3f(2) = – 3 and

e f ( x)
3x
4. sin 2 xdx equals
 x3 − 1 dx = − log | x − 1| + log | x + x + 1|
2

e3x m  2x + 1 
(1) (3 cos2x + 2sin2x) + C + tan −1   + C. Then (2m + n) is
13 n  3 
e3x (1) 3 (2) 5
(2) (3 sin2x – 2cos2x) + C
13 (3) 7 (4) 9

e3x
(3) (–3sin2x – 2cos2x) + C 2 x2 + 3  x −1 
13 8. If  ( x 2 − 1)( x 2 + 4)dx = a ln  x + 1  +
e3x
(4) (3sin2x + 2cos2x) + C x
13 b tan −1 + C , then values of a and b are
2
respectively.
x2 + 1 1 1
5.  ( x − 1)2 ( x + 3) dx equals (1) ,
2 3
(2) 1, 1

1 1 1 1
1 1 (3) , (4) ,
(1) e x −1 + . + tan −1 ( x + 3) + log x + C 2 2 3 3
4 ( x − 1)
3 1 1 1
x2 +
 2 1 
log | x − 1| − +C
e
(2)
 2 x − + 1 dx is
x
8 2 ( x − 1) 9. The value of integral
 x 
3 1 1 5 equal to (where C is the constant of integration)
(3) log | x − 1| − . + log | x + 3 | +C
8 2 ( x − 1) 8 x2 +
1 1
x 2 + +C
(1) e x +C (2) x 2
e 4
1 −1
(4) tan ( x + 1) + log}x + 3 | +C 2 1
x + 2
3 (3) xe x +C (4) x.e x + C
 2

tan 
10. The value of integral 11. e sec  – sin ) d equals
x  2 tan x  2  
 e  1 + tan x  + cot  x + 4  dx is equal to: (1) –etan sin + C
(2) etan sin + C
 
(1) e x tan  − x  + C (3) etan sec + C
4 
(4) etan cos + C
 
(2) e x tan  x −  + C
 4 x
 3  12. The graph of the antidervative of f(x) = xe 2 pass
(3) e x tan  − x  + C
 4  through (0, 3) then the value of g(2) – f(0) is
3  (1) 1 (2) 3

(4) e x tan  x −  + C (3) 5 (4) 7
 4 
 1

Lakshya JEE 2.0 (2024)

Indefinite Integration DPP-07

dx 2 x12  5 x9
1.  x5  x 6.  ( x5  x3  1)3 dx
1  1 
(1) log 1  4   c  x5 x10
4  x  (1) c (2) c
1  1  ( x5  x3  1)2 2( x5  x3  1)2
(2) log 1  4   c
2  x  x5  x10
(3) c (4) c
1  1  2( x5  x3  1) 2
2( x5  x3  1)2
(3)  log 1  4   c
2  x 
1  1 
(4)  log 1  4   c dx 1  xq 
4  x  7. Let  x2008  x  ln 
p  1  x r
  C , where p, q, r 

n
dx  1  N and need not be distinct then the value of (p + q
2.    1  m   c . Find mn.

x2 1  x 
5 4/5  x  + r) equals:
(1) 6024 (2) 6022
(3) 6021 (4) 6020
dx
3. The value of 
 
3/4
x2 x4  1
5 x 4  4 x5
8.  dx
x 
1/4
 x4  1  2
 x  1  x 1
1/4 5
(1)  4  c (2) 4
c
 x 
 
x5
 x 1
1/4 (1) x5 + x + 1 + C (2) C
 
1/4 4
(3)  x 4  1  c (4)   4  c x5  x 2  1
 x 
 
x5
(3) x–4 + x– 5 + C (4) C
dx x5  x  1
4.  ( x  2)7/8 ( x  3)9/8 
8 x 2
1/8
5 x  2
1/8 ( x  x3 )1/3
(1)    c (2)   c 9.  x4
dx is equals to
5 x 3 8 x 3
4/3
5 x 3
1/8
8 x 3
1/8 3 1 
(3)    c (4)   c (1)   1 C
8 x 2 5 x  2 8 x 2

3 
4/3 
1 
x 7 (2)  1  2  C
5.  dx 8 
 x  
1  x2 
5
4/3
1 1 
1 1 1 1 (3)  1 2  C
(1)  2  c (2)  2 c 8 x 
4 ( x  1)4 6 ( x  1)4
4/3
1 1 1 1 1 1 
(3)  2  c (4)  2 c (4)   1 C
8 ( x  1) 4 12 ( x  1)4 8 x 2

 2

1  x2 dx x 2009
10.  1  x2  14. If the primitive of the function
(1  x 2 )1006
w.r.t x
1  x2  x4
 1 m
(1) cos ec 1  x    c 1  x2  n
 x is equal to 
   C then is equal to____
n 1 x 
2 m
 1
(2)  cosec 1  x    c (1) 1 (2) 2
 x
(3) 3 (4) 4
 1
(3) cos1  x    c
 x dx
 1
15. 
(4)  cos 1  x    c ( x  1) x 2  1
 x
x 1 x 1
(1) c (2)  c
x 1 x 1
 x2  1 x 1 x 1
11.  4 2 dx is equal to: (3)  c (4) c
 x  3x  1  tan  x  x 
1  1 x 1 x 1

 1
(1) tan 1  x    c dx
16. 
 x
1  x2  1  x2
 1
(2) cot 1  x    c  1  x2 
 x 1
(1) tan 1  c
 1 4  2x 
(3) log  x    c  
 x  1  x2 
1
  1  (2) tan 1  c
(4) log  tan 1  x     c 2 

2x 

  x 
1  1  x2 
(3) tan 1  c
x 1  
 x(1  xe x )2 dx equals
2 2x
12.  

1  1  x2 
1
C (4) tan 1  c
(1) 4  2x 
1  xe x  
 xe x 
(2) log 
 1  xe x   C x2
  17.  is equal to

1  xe x 
x 2
 3x  3  x 1
(3)  log  x 
C
1  xe x   
 1  xe  (1)
1
tan 1 
x

3  3  x  1 
(4) x(e x  1)  C  

2  x 
7 4 1/2 (2) tan 1  
13. x (1  x ) dx is equal to. 3  3  x  1 
 
x 4  1(2 x 4  1) x4  1  
(1) 6
 c (2) 6
c (3)
2
tan 1  
x
6x 6x 3  x  1 


2 x4  1
(3) c (4) None of these (4) None of these
6 x6
 1

Lakshya JEE 2.0 (2024)

Definite Integration DPP-01

p dx  f ( x) = tan x − tan 3 x + tan 5 x −

1. If 0 1 + 4 x 2 = 8 , then the value of 𝑝 is 6. If
/4
with

(1)
1
(2) −
1 0 x
4
, then  f ( x)dx is equal to:
4 2 0
3 1 (1) 1 (2) 0
(3) (4)
2 2 1 1
(3) (4)
4 2
1
1
2.  3 + 4 x dx = 1

x
2
0 7. cos x3dx
7 7 0
(1) ln   (2) 4ln  
 3  3 (1) 3 sin1 (2)
sin1
1 3
(3) ln(7 / 3) (4) ln (3/7)
4 1
(3) sin   (4) None of these
3
1 x4 +1
3. The value of  dx is  /4
0 x2 +1 sin 2 x dx
1 1
8.  sin 4 x + cos4 x
dx
(1) (3 − 4) (2) (3 + 4) 0
6 6  
1 1 (1) (2)
(3) (3 + 4) (4) (3 − 4) 4 2
6 6 (3)  (4) 2

4
dx
4.  dx
 /2 sinx − cosx
8x − x2
0

9. 0 (1 + sinxcosx ) dx
(1) 0 (2)
4 (1) 0 (2) 1
 (3) –1 (4) None of these
(3) (4) 
2
sin −1 x
1
ln 2
x e− x dx =
10.  dx
5.  0
x(1 − x)
0  
−1 1 e (1) (2)
(1) ln(2e) (2) ln   4 2
2 2 2
2 2
1 1 (3) (4)
(3) ln(2 / e) (4) ln(2e) 4 16
2 2
 2

11. If f (x) is monotonic differentiable function on [a, 2 dx

b f (b ) 13. The value of 1 x(1 + x4 ) is equal to
b], then  f ( x)dx +  f −1 ( x)dx =
a f (a) 1 17 1 17
(1) log (2) log
(1) bf (a) – af (b) 4 32 4 2
17 1 32
(2) bf (b) – af (a) (3) log (4) log
2 4 17
(3) f (a) + f (b)
(4) Cannot be found 14. If f(x) = x3 + 3x + 4 then the value of
1 4

 f ( x)dx +  f −1 ( x)dx equals:

d  esin x  −1
F ( x) =   ; x  0 .
0
12. Let
dx  x  17
(1) 4 (2)
4
3 sin x 3

4
If e dx = F (k ) − F (1) , then one of the 21 23
1 x (3) (4)
4 4
possible value of k, is
(1) 15 (2) 16
(3) 63 (4) 64
 1

Lakshya JEE 2.0 (2024)

Definite Integration DPP-02

4 a
1. The value of  x − 1 dx is: 7.  x x dx =
0 −a
5 a a2
(1) (2) 5 (1) (2)
4 3 3
(3) 4 (4) 1 a2
(3) (4) 0
2
2
2.  (sin x+ | sin x |)dx is equal to: /2
 2 − sin 
0 8. The value of  log   d  is
 2 + sin 
(1) 0 (2) 4 −/2
(3) 8 (4) 1 (1) 0 (2) 1
(3) 2 (4) None of these
3. If f (x) = x – [x], for every real x, where [x] is the
greatest integer less than or equal to x. then, /4
1 9. The value of  log ( sec − tan ) d is:
−1 f ( x) dx is equal to: − /4
(1) 1 (2) 2 
(3) 3 (4) 0 (1) 0 (2)
4
 
x x (3)  (4)
4.  1 + 4sin 2
2
− 4sin dx is equal to:
2
2
0
2 /2
(1)  − 4 (2)
3
−4−4 3
10. The value of  ( x3 + xcosx + tan5 x + 1) dx is
 −/2
(3) 4 3 − 4 (4) 4 3 − 4 − (1) 0 (2) 2
3
2
(3)  (4) None of these
5. The value of 0 [ x2 ] dx , where [.] is the greatest
a

 ( sin )
integer function, is:
11. If 5
x + tan3 x + cosx dx , where , , 
(1) 2 − 2 (2) 2 + 2 −a
(3) 2 −1 (4) 2 −2 are constants, then the value depends on
(1)  ,a (2) , , ,a
2 x 2 + 1, x  1 2 (3) , ,a (4) 
6. If f ( x) =  3 , then  f ( x) dx is
 4 x − 1, x  1 0 1
ln  
 x2 
f   ( f ( x) + f (− x))
  3 
equal to:  
(1) 47/3 (2) 50/3
12.  g (3x 2 )( g ( x) − g (− x))
dx = .........
ln 
(3) 1/3 (4) 47/2
Where   1 .
(1) 0 (2) 1
(3)  (4) 1 / 
 2

 /3 1
1 x 7 − 3 x5 + 7 x3 − x
13.  tan x dx + 
−1
tan xdx = a − b + ln 2 14. The value of the integral  −1 cos2 x
dx
 /6 1/ 3
is:
then a / b is

3 3 (1) (2) 0
(1) (2) 2
2 2 
(3) (4) 2
3 3 4
(3) 3 3 (4)
2
 1

Lakshya JEE 2.0 (2024)

Definite Integration DPP-03

/2
7. Value of the integral I = 10 x(1 − x)n dx =
1.  sin 2 x dx
1 1 1
0 (1) (2) −
  n+2 n +1 n + 2
(1) (2)
4 2 1 1 1
(3) + (4)
(3)  (4) 0 n +1 n + 2 n +1

/2  a + b sin x  8. If [x] stands for the greatest integer function, then
2. 0 log   dx =
 a + b cos x  10  x2  dx
 
(1) 0 (2)  / 4   x2 − 28x + 196 +  x2  is
4    
(3)  / 2 (4) log / 2
(1) 1 (2) 2
7 x (3) 3 (4) 4
3. 2 x + 9− x
dx is equal to
9. If f and g are continuous function satisfying
1 3
(1) (2) f ( x ) = f ( a − x ) and g ( x ) + g ( a − x ) = 2 then
2 2
 0 f ( x ) g ( x ) dx =
a
5
(3) (4) 0
2
0 f ( x ) dx
a
(1) 0 (2)
 ecosx
4. 0 ecosx + e−cosx dx is equal to (3) a f ( x ) dx
0
(4) 1

(1)  (2)
2 3 dx
(4) − −
(1 + ex )(1 + x2 )
(3) 0 10.
3

/2 sin 2 x 
5. 0 sinx + cosx
dx = (1)  (2)
3
(1) 2log ( 2 +1 ) (2)
1
2
log ( )
2 +1 (3)

6
(4)

2

(3) log ( 2 +1 ) (4)

1
log ( )
2 −1  xdx
2 11.  01 + sinx =
 
6. 0 xf ( sinx ) dx = (1)
6
(2) 

  
(1) 0 f (sinx ) dx (2)  f ( sinx ) dx
0
(3)
3
(4) None of these

 
f ( sinx ) dx (4) None of these
2 0
(3)
 1

Lakshya JEE 2.0 (2024)

Definite Integration DPP-04


sin8 xlog cotx ( ) dx = /4
1.  2
0
7.  log ( cos x + sin )dx =
cos2 x
−/4
(1) 0 (2) 1
(1) π log 2 (2) − π log 2
1 
(3) (4) 
2 4 (3) − log 2 (4) π2 log 2
4

5
−2 cot ( tan x ) dx is equal to
−1  /2

2. The value of 8. sin 2 x cos2 x(sin x + cos x) dx =
− /2
7 7 2 2 4
(1) (2) (1) (2)
2 2 15 15
3
(3) (4) 2 6 8
2 (3) (4)
15 15
 dx
3. 0 1 + 2sin 2 x is equal to
9.
b
If f(a + b – x) = f(x), then  xf ( x)dx equal to
  a
(1) (2) b
3 3 3 a+b
2 a
(1) f (b − x)dx

(3) (4) 0
b
3 a+b
2 a
(2) f ( x)dx

4. Prove that b
  (3) b − a  f ( x)dx
2 2
− 2
 log ( sin x ) dx =  log ( cos x ) dx =
a
log 2
b
2 a+b
2 a
0 0 (4) f (a + b + x)dx

/2
| x | dx
5.  has the value /2n dx
−/2 8cos 2 x + 1
2
10. The value of 0 1 + tan n (nx)
is equal to, (n  N)
 2
 2
(1) (2) 
6 12 (1) 0 (2)
4n
2 2
(3) (4)  
24 18 (3) (4)
2n 2

ln x
 1 + x2 dx
51 dx
6. 11. Given f(x)(f(– x) = 9 then −51 3 + f ( x) has the value
0
(1) π (2) 2 π equal to
(3) 1 (4) 0 (1) 17 (2) 34
(3) 102 (4) 0
 2

 100
12. The value of 
0
3 log (1 + )
3 tan x dx is equal to 14. If 0 (1 − cos2x ) dx then the value of I is
 (1) 100 2 (2) 200 2
(1)  log2 (2) log 2
2 (3) 50 2 (4) 25 2
 
(3) log 2 (4) log 2
3 4 2
15. The solution of the integral 0 sin 4 x·cos6 x dx, is
13. The value of definite integral 3
 2
(1)
x 128
 dx,

− 1 + sin x + 1 + sin x
2
(2)
128
32 (2) 
3
(1) 5
2 3 (3)
128
2 (4) 
3 3
(3) 7
3 6 (4)
128
 1

Lakshya JEE 2.0 (2024)

Definite Integration DPP-05

2 5. The value of
ex sec x sin x
1. f ( x ) = cos x tan 2 x x then value of lim
(12 + 22 + ...n2 )(13 + 23 + .... + n3 )(14 + 24 + .... + n4 )

(15 + 25 + .....n5 )
n→ 2
3 4 x + tan x
is equals to:
/2

 ( x2 + 1)( f ( x) + f ''( x))dx is (1)

4
5
(2)
3
5
−/2
(1) 0 (2) 1 2 1
(3) (4)
(3) –1 (4) 2 5 5

4n
n 6. The value of
2. lim  is equal to
( )
1/ n
n→ r =1 r 3 r +4 n
2
 1   22   n 2  
lim 1 + 2  1 + 2  ....1 + 2   is
1 1
n→ 
 n   n   n  
(1) (2)
35 14 e/2
1 1 (1) 2
(2) 2e2 e/2
(3) (4) 2e
10 5
2
(3) e/ 2 (4) 0
e2
n n
+ 1

 ( n2 + i 2 ) n , then ln L is equal to :
3. Let Sn = 2n
( n + 1)( n + 2) ( n + 2)( n + 4) 7. If L = lim
1
n→ n 4
i =1
n 1
+ + …… + then lim Sn has 
(n + 3)(n + 6) 6 n' n→ (1) ln 2 + −2
2
the value of equal to:
(2) 2 tan–1 2 – 4 + 2 ln 5
3 9
(1) ln (2) ln (3) 2 tan– 1 2 + 4 + 2 ln 5
2 2
(4) 2 ln 5 – 4 – 2 tan–1 2
3 1 3
(3) 2ln (4) ln
2 2 2 1
 2 3 n
1
   
lim (1 + n ) n 1 +  1 +  ....2  is equal to

4. Let  > 1 and  > –1, then the value of n n n n
8.
n→   2  3 
 1 + 2 + .... + n   
lim n−    is :
n→  1 + 2 + ... + n
  1
 +1  +1 (1) e (2) e2
(1) (2)
 +1  +1 1 −1

+2 +2 (3) e4 (4) e2

(3) (4)
+2 +2
Evaluate the following limits
 1 1 1  
9. lim  + + + ... + 
n→  n 2 n2 + n n 2 + 2n n2 + n2 

 2

 1 1 1 1 3 n n n n 
10. lim  + + + ... +  13. lim 1 + + + + ... + 
n→ 1 + n 2 + n 3 + n 5n  n→ n  n+3 n+6 n+9 n + 3(n − 1) 

1  3  2 n 
11. lim sin + 2sin3 + ....... + n sin 3 
n→ n 
2 4n 4n 4n 

n −1
1
12. lim
n→

r =0 n − r2
2
 1

Lakshya JEE 2.0 (2024)

Definite Integration DPP-06

cos−1 ( t ) dt
cos x x4
2
2 t f (t )dt =3
3
1. The value of lim 1 8. If f (x) is a continuous function and
x→0 2 x − sin 2 x x
1 sin 2x, then f (1) is equal to
(1) 0 (2)
2 (1) 1 (2) –1
(3) −
1
(4)
2 (3) 3 (4) – 
2 3
x y
 dy 
x 9. If  (3 − x sin 2 t )dt +  cos t dt = 0 then,  
The points of extremum of  ( x ) =  e−t (1 − t 2 ) dt  dx ( ,)
2
/2 
2. 0
2
1
are: is equal to
(1) x = 1, −1 (1) – 3
(2) x = −1, 2 (2) 0
(3) x = 2, 1 (3) 3
(4) x = −2, 1 (4) None of these

x2  

1
dt t 2 f (t ) dt = 1 − sin x , x   0, 
3. If f ( x) =  1+ t
, x > 0 find f(9) 10. If
sin x
 2
then
2
 1 
f   equal to
t  3
x sin 2  
4. Find lim
f '( x)
where f ( x) =   2  dt 1
x →0 x t (1) 3 (2)
0 3
1
e
(3) (4) 3
1  1 3
5. Find  sin  x −  dx
1/ e
x  x
2

11. The solution of the integral 0 sin 4 x·cos6 x dx, is
6. If I n =  x n sin xdx find I5 + 20 I3. 3 
(1) (2)
0 128 128
5 7
x (3) (4)
ln t 128 128
7. Let f(x) =  1 + t dt , for x > 0, then the value of
1
If F(x) = 12
x
4 {4t − 2 F '(t )}dt , then F(4) equals
2
f(e) + f  1  is
12.
x
e
(1) 32/9
1
(1) 1 (2) (2) 64
2
(3) 64/9
1
(3) (4) None of these (4) None of these
4
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x t2 1 + u4
13. If F(x) = 1 f (t )dt , where f(t) = 1 u
du,

then the value of F (2) equals

7 15
(1) (2)
4 17 17

(3) 257 (4) 15 17

68

14. Let f: (0, ) → R be a continuous function such that

p f ( x) = 0 tf (t ) dt
x
. If F(x2) = x4 + x5, then
12
 f (r 2 ) is equal to
r =1
(1) 216 (2) 219
(3) 222 (4) 225