Indefinite Integration FOR IIT-JEE

INDEFINITE INTEGRATION L E V E L - 1 (Objective)

1.

If f(x) =

∫

2 sin x − sin 2 x

(A) 0 2.

If

∫

(B) 1 1 + sin

(A) 2 2 3.

4.

5.

If y =

∫

(

(D) 4 2

)

(A) A = - 1/4 & B may have any value

(B) A = - 1/8 & B may have any value

(C) A = - 1/2 & B = - 1/4

(D) none of these

∫

dx x  = I tan-1  m tan  + C then : 5 + 4 cos x  2

(B) m = 3

Given (a > 0) ,

∫

∫

(D) m = 2/3

(B) x > e

( ) dx is equal to :

(C) all x ∈ R

(D) no real x .

cot −1 e x e

(C)

∫

(C) I = 1/3

1 dx = loge a loge (loge x) is true for : x log a x

x

( ) +x+c

cot −1 e x 1 2x (A) ln (e + 1) 2 ex −1 x cot e 1

8.

(D) not defined

3/ 2 and y = 0 when x = 0, then value of y when x = 1 is : 1 + x2 1 2 (A) (B) 2 (C) 3 2 (D) 3 2 cos 4 x + 1 If ∫ dx = A cos 4x + B where A & B are constants, then : cot x − tan x

(A) x > 1 7.

(C) 2

x  x π dx = A sin  −  then value of A is : 2  4 4 1 (B) 2 (C) 2 dx

(A) I = 2/3 6.

where x ≠ 0 then Limit f ′ (x) has the value ; x→0

x3

( ) -x+c

2x

2

tan tan

ln (e + 1) -

−1 −1

e

x − cot

−1

x + cot

−1

x x

x

( )

cot −1 e x 1 2x (B) ln (e + 1) + +x+c 2 ex

(D)

1 2

2x

ln (e + 1) +

( ) -x+c

cot −1 e x e

x

dx is equal to :

(A)

4 2 x tan-1 x + ln (1 + x2) - x + c π π

(B)

4 2 x tan-1 x ln (1 + x2) + x + c π π

(C)

4 2 x tan-1 x + ln (1 + x2) + x + c π π

(D)

4 2 x tan -1 x ln (1 + x 2) - x + c π π

9.

If

∫

x4 + 1

(

dx = A ln |x| +

)

x x2 + 1

2

(A) A = 1 ; B = - 1 10.

ln | x|

∫

12.

(B) A = - 1 ; B = 1

1 1 + ln x (ln | x | - 2) + c 3 sin 2 x

2 arctan 2

(C) x -

2 arctan

∫

(D) A = - 1 ; B = - 1

(

(

(B)

2 1 + ln x (ln | x | + 2) + c 3

(D) 2 1 + ln x (3 ln | x | - 2) + c

w.r.t. x is :

1 + sin 2 x

(A) x -

∫

(C) A = 1 ; B = 1

dx equals :

Antiderivative of

)

2 tan x + c

(B) x -

)

(D) x -

2 tan x + c

 tan x   +c  2  2  tan x   +c 2 arctan   2 

1

arctan 

sin x . cos x . cos 2x . cos 4x . cos 8x . cos 16 x dx equals :

(A) 13.

+ c , where c is the constant of integration then 1 + x2

x 1 + l n |x| 2 1 + ln x (ln | x | - 2) + c (A) 3

(C) 11.

B

sin 16 x +c 1024

(B) -

cos 32 x +c 1024

(C)

cos 32 x +c 1096

(D) -

cos 32 x +c 1096

x 2 + cos 2 x cosec2 x dx is equal to : 2 1+ x

(A) cot x - cot -1 x + c cos ec x

(C) - tan -1 x - sec x

(B) c - cot x + cot -1 x +c

(D) - e ln tan

−1 x

- cot x + c

where 'c' is constant of integration . 14.

15.

16.

∫

3e x + 5e− x dx = Ax + B ln | 4e2x - 5 | + c then : 4 ex − 5e−x

(A)

A = -1, B = -7/8; C = const. of integration

(B)

A = 1, B = 7/8; C = const. of integration

(C)

A = -1/8, B = 7/8 ; C = const. of integration

(D)

A = -1, B = 7/8 ; C = const. of integration

x −1 1 . dx equals : x + 1 x2 1 x2 − 1 (A) sin -1 + x x 2 x −1 (C) sec -1 x +c x dx

∫

∫

x − x2

(B)

1 x2 − 1 + cos -1 + c x x

(D) tan -1

x2 +1 -

x2 −1 +c x

equals :

(A) 2 sin -1 x + c

(B) sin -1 (2x - 1) + c

(C) c - 2 cos -1 (2x - 1)

(D) cos -1 2 x − x 2 + c

17.

∫

when m, n ∈ N is equal to :

2mx . 3nx dx

) e( (B) +c m ln 2 + n ln 3 m ln 2 + n ln 3 x

2 mx + 3nx (A) +c m ln 2 + n ln 3

(C) 18.

19.

∫

2 mx . 3nx

(

ln 2 m . 3 n

)

+c

dx 3

cos x . sin 2x

2 (tan x)5/2 + 2 5

(C)

2 (tan2 x + 5) 5 dx

∫

m ln 2 + n ln 3

+c

equals :

(A)

If

(m n) . 2x . 3x

(D)

sin 3 x cos 5 x

tan x + c

2 (tan2 x + 5) 5

(B)

2 tan x + c

tan x + c

(D) none

= a cot x + b tan 3 x + c where c is an arbitrary constant of integration then

the values of ‘a’ and ‘b’ are respectively : (A) - 2 & 20.

If

∫

2 3

(B) 2 & -

22.

∫

(C) 2 &

(B) sin x

(D) cos 2x

cos 3 x + cos 5 x dx : sin2 x + sin4 x

(A) sin x - 6 tan-1 (sin x) + c

(B) sin x - 2 sin-1 x + c

(C) sin x - 2 (sin x) -1 - 6 tan-1 (sin x) + c

(D) sin x - 2 (sin x)-1 + 5 tan-1 (sin x) + c

∫

ln (tan x) dx equal : sin x cos x

1 2 ln (cot x) + c 2 1 2 ln (sin x sec x) + c (C) 2

24.

(D) none

(C) cos x

(A)

23.

2 3

eu . sin 2x dx can be found in terms of known functions of x then u can be :

(A) x 21.

2 3

π 4 1 π  (A) c - cot  2 x +  2  4 1 (C) (tan 4x - sec 4x) + c 2

∫

(B)

1 2 ln (sec x) + c 2

(D)

1 2 ln (cos x cosec x) + c 2

(B)

1 π  tan  2 x −  + c 2  4

 

sec2  2 x −  dx equals :

Primitive of (A)

x

(

3 x4 − 1

)

x4 + x + 1

x4 + x + 1

+c

2

(B) -

(D) none w.r.t. x is : x x4 + x + 1

+c

(C)

x +1 x + x +1 4

+c

(D) -

x +1 x + x +1 4

+c

L E V E L - 2 (Subjective)

cos 5x + cos 4x dx 1 − 2 cos 3x

1.

∫

4.

dx ∫ (1 − sincotx)x(sec x + 1)

2. ∫ cos x . ex. x2 dx

7. ∫ tan x . tan 2x . tan 3x dx 10. ∫ 13. ∫ 16. ∫

(

ln cos x + cos 2x 2

sin x dx

(x

2

17.

25.

[

31. ∫

3

(x + 2)

(

ex 2 − x2

∫

28. ∫

dx

(x − 1)

)

(1 − x ) 1 − x

3

x

∫

dx

19. ∫ x + x 2 + 2 dx 22. ∫

11.

14.

)

+ x +1

∫

(a > b)

5 x 4 + 4 x5 5

) dx

8.

2

(a + b cos x)

2

5

cos ec x − cot x . cos ec x + cot x

5. ∫

20. ∫

1/ 4

23. ∫

dx

26.

]

29. ∫

dx

32. ∫

3

(1 + x)

34. Integrate

3

1 f ′ (x) 2

1 + 2 sec x

sin x +

∫

cos 2 x sin x

∫

dx

(

cos x

15.

2

]

 x 2 + 1 l n x 2 + 1 − 2 ln x      x4  

(x

dx 3

)

+ 3 x 2 + 3x + 1 x

2

2 − 3x 2 + 3x

x

dx

x 2 + 2x − 3

dx

1+ x dx 1− x

2 − x − x2 2

3/ 2

( x sin x + cos x )

cot x −

∫

dx

w.r.t. x4 , where f (x) = tan -1x + ln 1+ x - ln 1− x

tan x

2

dx

dx

1 + 3 sin 2 x

18. ∫ ex

)

2

cos x

dx

[ (

x

3 12. ∫ e sin x . x cos x 2− sin x dx

dx

)

x −1 4

∫

9.

sin x sin (2 x + α ) sin x

dx

∫ sin x + sec x

dx 6.

dx

∫ (7 x − 10 − x )

1 − x dx 1+ x x

sec x

sin (x − a ) dx sin (x + a )

∫

3.

(x

2

)

+1

( x + 1) 2

dx





21. ∫ l n (ln x) +  dx (ln x) 2   1

(ax 2 − b) dx

24.

∫

27.

∫ (x

2

30.

∫ (x

2

33.

∫

x c2 x 2 − (ax 2 + b) 2 x ln x

)

−1

3/ 2

dx

x+2 + 3x + 3

)

dx x +1

dx ( x − α ) ( x − α ) ( x − β)

L E V E L - 3 (Questions asked from previous Engineering Exams)

1.

Find the indefinite integral



∫  (x) 

3x + 1 3 . (x + 1)

2.

Evaluate

∫ (x − 1)

3.

Evaluate

∫

f (x) x3 −1

1/ 3

[

]

l n 1 + ( x )1 / 6  1  + + ( x )1/ 4 ( x )1 / 3 + ( x )1 / 2  

dx

dx .

dx ; where f(x) is a polynomial of second degree in x such that

f (0) = f (1) = 3 f (2) = - 3 .

∫ cos 2 θ . ln

4.

Evaluate ,

5.

Evaluate

6.

Integrate ,

∫

7.

Let f (x) =

∫

∫

cos 2 x . sin 4 x cos 4 x . (1 + cos 2 2 x)

(

x3 + 3x + 2

)

Evaluate ,

∫

2

x 2 + 1 (x + 1)

dθ .

dx .

dx .

e x (x - 1) (x - 2) d x then f decreases in the interval :

(A) (- ∞ , 2) 8.

cos θ + sin θ cos θ − sin θ

(B) (- 2, - 1) 

  dx .  2  4 x + 8 x + 13 

sin -1 

2x + 2

(C) (1, 2)

(D) (2, ∞ )

ANSWER KEY LEVEL - 1 (Objective questions) 1.

b

13.

b,c,d

2.

d

14.

d

3.

d

15.

c

4.

b

16.

a,b,d

5.

a,b

17.

b,c

6.

b

18.

b

7.

c

19.

a

8.

d

20.

c

9.

c

21.

c

10.

a

22.

a,c,d

11.

a

23.

a,b,c

12.

b

24.

b

ANSWER KEY LEVEL - 2 (Subjective Questions) 1.

− (sin x + 1 x e 2

2.

sin 2x )+ c 2

[(x

2

)

− 1 cos x + ( x − 1) 2 . sin x

] +c

(

)

3.

  cos a . arc cos  cos x  - sin a . ln sin x + sin 2 x − sin 2 a + c

4.

1 1 x x x ln tan + sec² + tan + c 4 2 2 2 2

5.

x 1 sin-1  sec 2  + c

6.

1 ln 2 3

7.

1 1   − ln (sec x) − 2 l n (sec 2x) + 3 ln (sec 3x) + c  

8.

-

9.

sin x − x cos x +c x sin x + cos x

 cos a 

2

2

3 + sin x − cos x + arc tan (sin x + cos x) + c 3 − sin x + cos x

[

]

1 2 ln cot x + cot α + cot x + 2 cot α cot x − 1 + c sin α

((

cos 2x - x - cot x . ln e cos x + cos 2x sin x

10.

)) + c

11.

1 t2 − 2 t + 1 1 1 4 ln (1 + t) - ln (1 + t ) + 2 2 ln 2 - tan-1 t2 + c where t = 4 t + 2t +1 2

12.

esinx (x - secx) + c

13.

14.

−

-

b sin x 2a + 2 a − b (a + b cos x) a − b2

(

2

2

)

 1  ln 2 2 

(

2 cot x − cot 2 x − 1 2 cot x + cot 2 x − 1

)

3/2

+ ln

arc tan

a−b x . tan a+b 2+c 

( cot x − 1 + cot x) + c 2

cot x



2 sin 2 x   +c   sin x + cos x 

15.

tan-1 

17.

3 x  x − 1 3 tan-1 x ln   +c 4 16  x + 1 4 x −1 8

19.

1 3

21.

xln (lnx) -

(

x+

x2 + 2

3/ 2

x lnx

8 (x + 1) 2

+

(

2

- x + x2 + 2

)

1/ 2

+c

ex 1 − x + c

27

arcsec x −

28.

ln

29.

8 3

x2 −1

u +u +1 4

2

18.

ex x + 1 + c

5

x −1

(x

2

)

+1

x2 +1  1   . 2 − 3 ln 1 + 2   3  9x x  

22.

4  x −1    3  x + 2

 2  1 . cos-1  x + 1 + c 16

24.

 ax 2 + b  sin −1   +k  cx 

9 7 x − 10 − x 2

4x

2

+ 3 tan −1

1 + 2u 2 3

+ c where u = 3

1− x 1+ x

)

x +c 3 ( x + 1)

1+ x

+

15 ln 8

1+ x −1 1+ x +1

+c

32.

 4 − x + 2 2 2 − x − x2  2 − x − x2 2  − sin −1  2x + 1 + c l n − +    3  x x 4  

33.

−2 . α −β

34.

- ln (1 - x4) + c

x −β +c x−α

+c

+c

(

15 x 2 + 5x − 2

+c

2 (7 x − 20)

 −1  5 t − 1  1   − sin −1 x − 1 − x 2 + c where t = ln   tan t + 2 5  5 t + 1  

2 arc tan 3

1/ 4

+c

26.

ln x

| u 2 − 1|

-

20.

1+ x

25.

31.

)

x 2 + 2x − 3

23.

30.

)

(

x+1 +c x + x+1

16.

1+ x 1− x

ANSWER KEY LEVEL - 3 (Questions asked from previous Engineering Exams) 1.

I = I1 + I2 + c , where ;  y8 8 y 7  28 y 6 56 y 5 70 y 4 56 y 3 28 y 2 I 1 = 12  − + − + − + − 8 y + 1 n y 7 6 5 4 3 2 8   1  1 I 2 = 2 e 3z  z −  − 9 e 2 z  z −  + 18 e z (z − 1) − 3 z 2 + c 2  3  2

; where z = ln (1 + x1/6)

1 x +1 x +1 ln +c − 4 x − 1 2( x − 1) 2

2.

3.

 x2 + x +1 2  2 x + 1 ln  x − 1  + 3 arctan  3  + c    

4.

(b)

5.

2 ln (1 + cos 2x) +

6.

x 3 1 1 tan -1 x - ln (1 + x) + ln (1 + x2) + +c 1 + x2 2 2 4

7.

C

8.

-

1  cos θ + sin θ  1 (sin 2 θ ) ln  ln (sec 2 θ ) + c 2  cos θ − sin θ  2

2 1 + cos 2 x

; where y = x 1/12 + 1

- ln (1 + cos2 2x) + c

2x + 2 1 3  2x + 2  tan −1 − log ( 4 x 2 + 8 x + 13)   3 2 2  3 