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INTEGRATIONS AND APPLICATIONS OF DEFINITE INTEGRALS

1. (d) Put sec x2 = t 2x sec x2 tan x2 dx = dt

x sin x2 esec x2 cos2 x2

dx = 1 2

et dt = 1 et + c = 1 esec x2 + C

2

2

2. (c)

Put tan x = t 2

Then

cos

x =

1 - t2 1 + t2

,

sin

x

=

2t 1 + t2

dx =

2dt 1 + t2

2dt

dx

=

3 cos x - 4 sin x + 5

1 + t2

3

1 1

- +

t2 t2

-

4

?

1

2t + t2

+5

=

2dt 3(1 - t2) - 8t + 5(1 + t2)

=

dt t2 - 4t + 2

=

dt (t - 2)2

= -

1 t-2

+

C

=

-

1 tan x

-2

+ C

2

=

1 2 - tan x

+

c

2

3. (a) Put x = t2 dx = 2t dt

cos x dx = cos t. 2t dt = 2 t cos t dt

= 2[t sin t - sin t dt] = 2 [t sin t + cos t] +c

= 2 [ x sin x + cos x ] + c

4. (b)

sin x dx sin2 x + 4 cos2 x

=

sin x dx 1 + 3 cos2 x

Put 3 cos x = t - 3 sin x dx = dt

=

- 1 dt

3 1 + t2

=

1 cot ? 1 t + c = 3

1 cot ? 1 ( 3

3 cos x) + c =

1 3

sec x tan ? 1 3 + c

5. (c) Put ex = t, then ex dx = dt

ex dx

=

(1 + ex ) (2 + ex )

dt

=

(1 + t) (2 + t)

1 - 1 dt 1 + t 2 + t

= log (1 + t) ? log (2 + t) + c = log 1 + t + c = log 1 + ex + c

2+t

2 + ex

6. (d)

( ) tan x + cot x dx =

sin x + cos x dx = 2 sin x cos x

(sin x + cos x) dx 1 - (sin x - cos x)2

Put sin x ? cos x = t

(cos x + sin x)dx = dt

2

dt = 2 sin ? 1 t + c = 1 - t2

2 sin ? 1 (sin x ? cos x) + c

7. (b) Put 1 + x2 = t, then 2x dx = dt

 x3

dx =

(1 + x2 )2

x2 x dx = (1 + x2 )2

(t - 1) dt t2 2

= 1 2

1 t

-

1 t2

dt

=

1 2

log

t+

1 t

+

c

=

1 2

log (1 + x2) +

1 2 (1 + x2 )

+ c

8. (a)

tan x sec x dx =

tan x sec x dx = 2 sec x + c sec x

9. (b)

ex

x+4 (x + 5)2

dx

=

ex

x +5-1 (x + 5)2

dx

=

ex

x

1 +

5

-

(x

1 + 5)2

dx

=

ex (f (x) + f | (x)) = ex f (x) = ex

x+5

10. (a)

4x2

dx + 12x

+

45

=

1 4

dx x2 + 3x + 45

4

=

1 4

dx x + 3 2 + 45 - 9

2 4 4

= 1 4

dx x + 3 + 32

2

= 1 4

?

1

tan ? 1

x + 3 2

=

1 tan ? 1 2x + 3 + c

3

3

12

6

11. (d)

e 2log tan sin x cos x dx =

elog tan2 sin x cos x dx

[Q e log t = t]

= tan2 (sin x) cos x dx = (sec2 (sin x) ? 1).cos x dx = (sec2 t ? 1) dt

= tant ? t = tan sin x ? sin x + c 12. (b)

ex (sin2 x + 2 sin x cos x) dx = ex (f (x) + f|(x)) = ex f(x) = ex sin2 x

13. (a)

sin3 x

+

cos3 x

=

sin s2x cos2 x sin2 x cos2 x

sin x +

cos2 x

cos x sin2 x

= (sec x tan x + cosec x cot x)dx = sec x ? cosec x + c

14. (d)

1 - sin2x =

cos2 x + sin2 x - 2 sin x cos x

=

(cos x - sin x)2dx = cos x ? sin x dx = sin x + cos x

15. (b)

ex

1+ 1+

sin cos

x x

dx

=

1 + 2 sin x cos x

ex

2 2 cos2 x

2

2 dx

=

ex 1 sec2 x + tan x dx

2

2

2

= ex tan x + c 2

This is in the form ex (f (x) + f | (x)) dx Answer is ex f(x) i.e. ex tan x/2

16. (d)

1

- cos2 cos4 x

x

dx

=

sin2 x cos4 x

dx =

tan2 x sec2 x dx = tan3 x 3

+ c (Put tan x = t)

17. (a)

log 5

0

ex ex - 1 ex + 3

dx

Put ex ? 1 = t2 ex dx = 2t dt when x = log 5, t = 2 x = 0, t = 0

log5 0

ex ex - 1 ex + 3

dx =

2 0

2t2 4 + t2

dt

=

2

2 0

t2 + 4 - 4 + t2

4

dt

=

2

2 0

= 2

t

-

2

tan- 1

t 2

2 0

= 2

2

-

2

4

= 2

4 - 2

= 4 -

18. (a)

tan x 1/e

t dt 1 + t2

+

cot x 1/e

dt t (1 + t2)

1 -

4

4 + t2

dt

Put t = 1 u

dt

=

-

1 u2

du

When t = cot x, u = tan x and t = 1 , u = e e

=

tan x 1/e

t dt 1 + t2

+

tan x e

-

1 u2

du

=

1 u

1

+

1 u2

tan x 1/e

t dt

1 + t2

-

tan x e

=

tan x 1/e

t dt 1 + t2

+

e tan x

t 1 + t2

dt

=

e 1/e

1

t + t2

dt

u 1 + u2

du

( ) = 1 log 1 + t2 2

e =1 1/e 2

log (1

+

e2)

-

log

e2 + e2

1

=

1 2

log

(1 + e2) (e2 + 1)

=

1 2

log e2 = loge e = 1

e2

19. (b)

Put loge x = t, then

dx = dt x

When x = e37, t = loge e37 = 37

X = 1, ,t = loge 1 = 0

e37 1

sin ( loge x) dx = x

37 0

sin

t

dt

=

-

cos t

37 0

= - [cos 37 - cos 0] = - [-1 ? 1] = 2

20. (c) Put xex = t, then (1 + x) ex dx = dt

ex (1 + x) sin2 (xex)

dx

=

dt sin2 t

=

cosec2 t dt = - cot t + c = - cot (xex) + c

21. (c) (1 + cos x + cos2 x + ......... to ) =

1

-

1 cos x

dx

(Q

it

is

a

G.

P)

=

1 2 sin2 x

dx =

1 2

cosec2

x 2

dx = - cot

x 2

+ c

2

22. (a)

sec x tan x 10 + tan2 x

dx

=

sec x tan x dx 9 + sec2 x

=

dt 32 + t2

(Put sec x = t, sec x tan x dx = dt)

= 1 tan ? 1 t + c = 1 tan ? 1 sec x + c

3

3

3

3

23. (d)

dx a2 cos2 x + b2 sin2 x

=

sec2 x dx a2 + b2 tan 2 x

= 1 b

dt a2 + t2

[Put b tan x = t, b sec2 x = dt]

= 1 ? 1 tan ? 1 t + c = 1 tan ? 1 b tan x + c

ba

a

ab

a

24. (b)

x2

1

+

1 x2

x2

x2

+

1 x2

dx

=

1

+

1 x2

dx

x

-

1 2

+

2

x

Put x -

1 x

=

t,

then

1 +

1 x2

dx = dt

=

dt t2 + 2

=

25. (b)

1 tan ? 1 2

t 2

+ c =

1

x - 1 tan ? 1 x + c =

2

2

1 2

tan ? 1

x2 - 1 2 x

+ c

1 + sin x dx 3

=

sin2 x + cos2 x + 2 sin x cos x

6

6

66

dx =

sin x + cos x 2 dx

6

6

= - 6 cos x + 6 sin x + c = 6 sin x - cos x + c

6

6

6

6

26. (a)

sec2 x cosec2 x dx =

cos2

1 x sin2

x

dx

=

sin2 x + cos2 x cos2 x sin2 x

dx

=

= (sec2 x + cosec2 x) dx = tan x ? cot x + c

27. (c) cosec4 x dx = (1 + cot2 x) cosec2 x dx

put cot x = t -cosec2 x dx = dt

=

(1 + t2) (-dt) = -

t

+

t3 3

+ c = - cot x -

cot3 x 3

+ c

28. (a)

1 cos2

x

+

1 sin2

x

dx

1-x 1+ x

dx =

1-x 1+ x

?

1 - x dx 1-x

=

1-x

dx =

1

-

x

dx = sin ? 1 x + 1 - x2 + c

1 - x2

1 - x2 1 - x2

29. (d)

sinh logx dx =

elog x - e- log x

2

dx =

x

-1 2

/

x

dx

=

?

x2

2

-

log

x

=

x2 - log 4

x+ c

30. (b)

Take 2x - 2 = t, 2x ?1 = 2x ? 2 + 1 = t2 + 1

2 dx = dt 2 2x - 2

dx = tdt

dx

(2x - 1) 2x - 2

=

tdt (t2 + 1)t

= tan ? 1 t = tan ? 1

2x - 2 + c

31. (a) xx loge ex dx = xx (loge e + loge x) dx = xx (1 + loge x) dx = xx + c

32. (a)

cos x cos x -

dx =

4

cos x dx 1 (cos x + sin x) 2

=

2 2

cos x - sin x + cos x + sin x dx = (cos x + sin x)

1 2

cos x

cos

x

- sin x + sin x

+

1

dx

= 1 [log (cos x + sin x) + x] + c 2

33. (a)

x -1 (x + 1)3

=

(x + 1) - 2 (x + 1)3

=

1 (1 + x)2

- 2 (x + 1)3

= f (x) + f | (x)

I=

ex [f (x) + f | (x)] dx = ex f (x) = ex 1 (1 + x)2

34. (a)

Nr = l (Dr) + m

d dx

(Dr)

Sin x + 3 cos x = l (3 sin x + 4 cos x) + m (3 cos x ? 4 sin x)

= (3l ? 4m) sin x+ (4l + 3m) cos x

3l ? 4m = 1

4l + 3m = 3

9l ? 12m = 3

16l + 12m = 12

25 l = 15

l= 3 5

m= 1 5

sin x + 3 cos x 3 sin x + 4 cos x

dx

=

3 5

x

+

1 5

log (3 sin x + 4 cos x)

A + B =

3+ 5

1 5

=

4 5

35. (d)

In + In ? 2 =

4

cotn x dx +

4

0

0

cotn ? 2 x dx =

4

0

(cotn x + cotn ? 2 x) dx

4

= 0

4

cotn ? 2 x (cot2 x + 1) dx = cotn ? 2 x . cosec2 x dx 0

=

-

cot n -1

x

4

n - 1 0

=

-

n

1 -

1

-

0

=

1 1-n

36. (d)

0

dx cos2 x + 3 sin2 x

=

0

sec2 x 1 + 3 tan2 x

sec2 x 1 + 3 tan2 x

is discontinuous at x =

2

37. (d)

2 2 + x dx = 2

0 2-x

0

2 + x ? 2 + x dx = 2

2-x 2+x

0

2 + x dx = 2

4 - x2

0

2 + 4 - x2

x 4 - x2

dx

=

2

sin- 1

x 2

-

4

-

x2

2 0

= [2 sin ? 1 1 ? 0] ? [0 -

4]=2? +2=+2 2

38. (b)

1 sin2 (cos ? 1 x) dx 0

= 1 1 ? [cos(cos ? 1 x)]2 dx = 1 1 ? x2 dx = x ? x3 1

0

0

3

0

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