UNIT # 06 INDEFINITE & DEFINITE INTEGRATION INDEFINITE INTEGRATION

EXERCISE - 01 1.

2 sin x  sin 2x , x0 x3 2 sin x  1  cos x  lim f '(x)  lim  1 x 0 x 0 x  x 2  f(x)  

8

4.

9.

 cos x  sin x  dx 2 x cos 2 x 

14.

 x(x

(x 4  1) dx =  1  2 x2 )

4

x



  (1  x

5

 ) 

3/2



  x

4

2x 1  dx  1  2x 2 x 

 

17.

x4  4

 x

 1 + c  dx = n|x|+ 2  x 1

x 2 1 1    dx 2 x  1 5 (x  1) 5(x  2)  2

1 2 1 n(x 2  1)  tan 1 x  n| x  2|  K 10 5 5



3

4  1  x2 x2

x dx (1  x 5 ) 3 / 2 20.

1 dt 2  (1  x 5 ) 1 / 2  C 5  (t) 3 / 2 5

1  2 tan x sec x  2 tan 2 x = |sec x + tanx|

I I I I

 2  n(x  1  x )dx 1x  II I

= = = =

n|sec x + tan x| + n|sec x|+ c n|sec 2 x + secx tan x| + c –n|secx – tanx|+ n|secx|+ c n|1 + tanx (sec x + tanx)| + c

=



(2 sin 2x cos x  6 sin 4x cos x  6 sin 6 x cos x) dx sin 2x  3 sin 4 x  3 sin 6x



2 cos x(sin 2x  3 sin 4 x  3 sin 6 x ) dx sin 2x  3 sin 4 x  3 sin 6 x

= 2sinx + c

.....(i)

BRAIN TEASERS 5.

I =

 sin

2

(  n x) dx ,

let t = nx  dt = =

dx

 | sec x  tan x| dx = I

2

(sin x  sin 3x)  3(sin 3x  sin 5 x)  3(sin 5 x  sin 7x) dx  sin 2x  3 sin 4 x  3 sin 6 x

4x 2  1  x 2

1 1 / 2 [put t = 4x –2 + 1 + x 2 ] t dt 2 4 4  x2  x 4  1  x2  C  C = 2 x x

EXERCISE-02 2.

x  4x 3

dx  

=

x

 

 1

  5



6

dx  

1 Ax  B C  2  2 (x  2)(x  1) x 1 x  2

Let

1 2 1 On solving it we get A   , B  , C  5 5 5

sin 2 x 2 2 C =  (cos x  sin x) dx   cos 2xdx = 2 x1 / 3 1 4 x 5 dx  dx  (x 4  1) 4 / 3 4  (1  x 4 ) 4 / 3 3  (1  x 4 ) 1 / 3 + C (Put 1 – x –4 = t) 4

{Let t = 1 + x –5; dt = – 5x –6dx}

13.

 1  x 2 n(x  1  x 2 )  x  C

(cos 4 x  sin 4 x)(cos 2 x  sin 2 x) dx (1  2 sin 2 x cos 2 x)

 2x = n|x|    2 2  (x  1)

10.

  1   1  x 2 n(x  1  x 2 )   1  x 2 . dx  2 1x  

8

  1  2 sin 

5.

CHECK YOUR GRASP



I =

I =

1 2

t

 e sin t

2

dx x

t dt

 e (1  cos 2t) dt

 dx = e t d t

 2I = e t – Let I1 =

t

 e cos 2t dt

14.

 cos

t

 e cos 2t dt

=

et cos 2t  2 sin 2t   C = 5

 

e

(

ax

I =



cos bxdx 

e

ax

a 2  b2

17.

=

  dx 

x c x4  x  1

 [x

6

dx x 6  dx 5 1 / 3 (1  x )] (1  x 5 )1 / 3



1 sin x dx = 4 sin 4 x



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

I =

1 = 4



c

=

1 4 1 4

 (1  t

2

dx cos x.cos 2 x

dt

 1 t

dt 1 2 dt= ) (1  2t ) 4

– 2

1 4

1 = 4



2

  1  2t

2



1   dt 1  t2 

1  t2 t

 2 tan 

d

dt

[put tan = t]

1  t5 / 2  2 2 t    5 / 2 2 5  dx

 (1 



x) x x

dx =

2

dx 1  x  1  x 

–

3/2

2

2 x

2

–

1x



(1  x )



x (1  x) 3 / 2



dx

dx

 (1  x)

3/2

2( x  1)

+ C =

1x

1x

f(x) =



tan  5  tan 2   c

+ C

1 1 – x . cos x x

  3x

2

.sin

1 1  x cos  dx x x

1 1 1  1  . –  cos   2  x 3 dx –  x cos dx x x x  x  1 +C x 

C = 0

1  3  x sin , x  0 f(x) =  x  0 , x 0

f(x) is clearly continuous and differentiable at x = 0 zero with f' (0) = 0. f"(0) = hlim 0

3h 2 sin

1 1  h cos h h h

dt

1t

2

 sin x 1  sin x 1 2 n – n +C 1  sin x 1 8 2  sin x 2

1

 2

= x3 sin



2

1

1

d 4

1  since f   = 0 + C 

Putti ng t = si nx, we get dt = cosx.dx I=

 cos

2 tan 

=x 3 s i n

2/3



sec 2 (1  tan 2 )

f'(x) = 3x 2 . si n 

3  x5  1  1 (1  x 5 ) 2 / 3 =  dx =   10  x 5  5 2/3 13.

=

18.

  f(x)  xf '(x)dx  xf(x)  C  12.

I=

1 t e (5 – 2si n2t – cos2 t) + C 10

 3x 4  1 1 x(4x 3  1)  (x 4  x  1)2 dx    x 4  x  1  (x 4  x  1)2

=

 sin 2 

a cos bx  b sin bx  )

x = (5 – 2si n(2nx) – cos(2nx)) + C 10

11.



d 3

= 3hsin

1 1 – cos h h

This limit does't exist, hence no n- di ff ere nt i a b l e a t x = 0.

f'(x)

is

Also lim f'(x) = 0. Thus f'(x) is continuous at x 0 x = 0.

EXERCISE - 03

MISCELLANEOUS TYPE QUESTIONS

Fill in the blanks : 1.

(C)

4e 2 x  6  9e 2 x  4 dx = Ax + Blog (9e 2x – 4) + C

Now differentiate both sides

4e 2 x  6 B(18e 2 x )  A  9e 2 x  4 9e 2 x  4

2.

3 35 ,B  ,C R 2 36

 (log (log x)  (log x)

1 a2

x

(D)

2

dx  C

x 2 dx+  (log x) dx + C x log x (integrating by parts the first term)

x a

1 2

x a

2

2

2

x -a



2



a

2

1 | x|  1  a  sec 1   cos 1    a  a  a | x| 

 1 1  a   sin  | x| 2 a

1 a  C  sin 1 a | x| Assertion & Reason :

+  (log x) 2 dx +C

1.

(again integrating by parts)

= x log (log x) – x(logx) –1 + C Putting x = e, we have 1998 – e = e.0 + e + C. Thus C = 1998 Match the column :

Let D(x) =  1 f 1(x) +  2f 2(x) +  3 f 3 (x) where  1 = (b 2c 3 – b 3c 2 ),  2 = a 2c 3 – a 3c 2 ,  3 = a 2 b 3 – a 3b 2 then

 D(x ) dx

=

 1 f1 (x ) dx

+

 2 f2 (x) dx

+   3 f3 (x) dx + C ... (1)

dx  (a  x 2 )3 / 2 , put x = a tan 2

 f1 (x)dx  f2 (x)dx  f3 (x)dx

2

x a+ 2

2

a sec  d  1  3  2 sin  a sec 3  a

a



2

x

a2  x2

C

 x 2  a 2  a2 dx       dx 2 2 2 2 2 2 a x a x   a x 2

   a  x dx  a

a2

b2

c2

a3

b3

c3

+ C

Thus statement-I is true and follows from statementII which we have applies at Eq. (1)

x2

2

=

 a

x



(B)

x

x 2

= xlog(logx) – [x(logx) –1 +  (log x) 2 dx]

(A)

Put x = a sec

 1   sin   + C  

c

= xlog(logx)– 

2.

a tan  sec  1 cos  d  2  2 d a 3 (tan 3 ) a sin 

a

An antiderivative of f(x)= F(x) =



c

on comparing we get

A

2



4e 2 x  6 9 Ae 2 x  18Be 2 x  4 A  2x  9e  4 9e 2 x  4

dx  a 2 )3 / 2

 (x

2



dx a2  x2



x 2 a2 x x a  x 2  sin 1    a 2 sin 1 2 2 a a  



x 2 a2 x a  x 2  sin 1    C 2 2 a

3.

The statement-II is false since in

dx

 x  3y

= log(x – 3y) + C, we are assuming that y is a constant. We will now prove the statement-I. x , and From the given relation (x – y)2 = y 2log(x – y) = logx – logy ... (1) Also,

dy  y  xy =   . . dx  x  x  3y

To prove the integral relation, it is sufficient to show

1 d that RHS. = x  3y dx

2.

x x   1 2 Now, RHS = log   1   (x  y )   y 2 y  

3.

 1  log x  log y  log y  2  2 

=

d 1  1 3 dy   RHS.=    dx 4  x y dx 

1 1 1  3   y  x  y      = x y  x  x  3y x  3y 4   Thus, statement-I is true. Comprehension # 2 : =

I n= =

ax 2  2bx  c

dx

xn a

ax 2  2bx  c –



x n  1 (ax 2  2bx  c) ax 2  bx  c

dx ...(ii)

(n + 1) a u n+1 + (2n + 1)bu n + ncu n–1 = x n ax 2  2bx  c

CONCEPTUAL SUBJECTIVE EXERCISE

1 1 1 1     2 1x 2  1  x 1  x 

tan 2  d =  cos 2 (tan   )

1 1 2   1  x2 1  x2 1  x 4

Put

4 x3  4  d(x )   1  x 4 dx 

= – n(1 – x 4) + C

I =

... (i)

x n (2ax  2b)



b u a n

from (i) and (ii) be get

x 1  2n 1 1 In = . 2 2 n 2 2 2n a 2na (x  a )



ax  2bx  c

dx –

–  nx n 1 2 ax 2  2bx  c dx

n a

x +2n(I n – a 2I n+1 ) (x  a 2 ) n

Put x = tan

2

1 n x 2 ax 2  bx  c 2a

=

EXERCISE - 04[A]

8.

b u , where a n

1 2a

dx

ax 2  2bx  c

= In –

1 dx (x  a 2 ) n 1

1 2 Now   2  1  x4

x n (2ax  2b)  2bx n

x n (2ax  2b)

2–m

dx

ax 2  2bx  c



=

2

f'(x) =

x n 1

1 = 2a

2

Whence I n+1 +





  x2  a2   x a2 +2n  dx    2 2 n  2 2 n 1 2 2 n 1  (x  a ) (x  a )     (x  a )

–2na 2 

5.

1 2a

In =

x2 x +2n  (x 2  a 2 ) n 1 dx (x 2  a 2 ) n

1 x dx = 2 +2n  2 2 n (x  a 2 ) n (x  a )

=



=

1 = [logx – 3logy] 4

1.

u n+1 =

[From Eq. (1)]

sin n  2 x  cos m  2 x dx

n 1 sin n 1 x – I m 1 m  1 n–2, (m  1) cos x

=

1 [log(x – y) – logy] 2

=

n 1 sin n 1 x – m 1 m 1 (m  1) cos x

In, –m =

I

du

 cos

2

u

dx = sec d

tan 2  sec 2  2

 sin    cos  sin(tan )  cos(tan )   

d



tan 2d = du

  sec 2 udu  tan(tan   )

 tan(x  tan 1 x)  2



tan     u

tan x  x 1  x tan x

sin x  x cos x C cos x  x sin x

9.

 cos   sin  

 cos 2n  cos   sin   d

I

sin 2  2  cos   sin   sin 2   n   . d  cos   sin   2 2 cos 2  

10.

= 2

  x  nx  x  dx    e  

t 

 2

x

3x 2  1 3x 2  1 dx   (x 2  1)3  (x  1)3 (x  1) 3 dx

I

1 1 1       dx 3 2  (x  1) (x  1) 3  

15.

16.

1  (x  1) 2 (x  1) 2  x    C 2 2  2 2  (x  1) 2 sec 2 x dx dt  [Put tan x = t] 2 x  2 tan x  t 2  2t

1 1 1  1 tan x  dt  n C    2 tan x  2 2  t t 2

 (sin x)

11 / 3

cos x 1 / 3 dx  

4



1 t

1

2

dt  2 

2 2

,

2/3

t

2

)dt [Put cot x = t]

8/3

I =  ( tan x  cot x )dx 



(sec 2x)dx = 2tdt

1 t

 tan x  cot x  2 tan 1    C  2   cos 2x dx sin x

cos 2 x  sin 2 x dx   cot 2 x  1 dx sin x

I =  sec 2   1.

 

 

sec . tan 2  1  sec 2 

tan x  1

sec . tan 

d  

1  cos 2  cos (1  cos 2 )

d

(1  sec 2 ) sin 2 

cos   cos 3 

d

d

(1  cos 2 )  2 cos 2  cos (1  cos 2 )

d

cos  1  cos 2 

  log| sec   tan | 2    log| sec   tan | 2 

dx

d cos 

2  sin 2  dt

d

(put sin = t)

2  t2

tan x

  log| sec   tan | 2.

put tanx = t 2 

 2

 u  tan 1  C 2  2



 

t  t 3    [4 cot 2 / 3 x  cot 8 / 3 x]  C C  2 / 3 8 / 3 8  

19.

dt 2

2

= –  sec d   2     (t

2

where u = t –

 2

1 t2

 1 t  t    

–cosec 2x dx = sec tand

1 dx (sin x cos x)1 / 3 2

5/3

dt



11

cos ec x (1  cot x) cos ec x dx   dx (cot x)1 / 3 (cot x)1 / 3 1 / 3

t4  1

Put cotx = sec

 tan 





t2

u 

x

I =

1

t2  1

dt  2 

t 1

2

2



20.

2t 4

du

=

1 1  x e I    1  2  dt  t   C        C t t   e x

.

t2

2

x x   x x Put    t  nx   dx  dt  e e  

13.

t2  1

1

1  cos   sin   1 n  sin 2   n(sec 2 )  C  2  cos   sin   2

   x  x  e  x  1 I =        nx dx    1  2x e x x       e   



 dx 

2t 1  t4

dt   log| cot x  cot2 x  1| 

1

log

2 2 1 2

log

2  sin  2  sin 

2  1  tan2 x 2  1  tan 2 x

C

C

EXERCISE - 04 [B]

BRAIN STORMING SUBJECTIVE EXERCISE

cot xdx cos 2 x dx   (1  sin x)(sec x  1)  sin x(1  sin x)(1  cos x)

4.



(1  sin x) dx 1 dx    dx sin x(1  cos x) sin x(1  cos x) 1  cos x 

sin x dx x  tan  C 2 (1  cos x)(1  cos x) 2



 dt x  tan  C 2 (1  t)(1  t ) 2



13.

x

1x C 1x

f(x )dx x   1   dx (x  1) 3   x 2 (x  1) 3 

2

f(x)dx 3x 2  3x  1   x 2 (x  1)3  x 2 (x  1)3 dx  f(x) = 3x 2 + 3x + 1 f'(0) = 3

1 x 1 x x n tan  sec 2  tan  C 2 2 4 2 2

I

8.

 ex

14.

e cos x (x sin 3 x  cos x ) dx  sin 2 x

put cos x = t, we get –sinx dx = dt

n(cos x  cos 2x ) dx sin 2 x

  cos ec 2 x n  (sin x)(cot x  cot 2 x  1 )  dx  

dx 

  n(sin x ) cosec x dx

 dt 1  t2

2

e

  n(cot x  cot 2 x  1 ).cosec 2 x dx  n(sin x) cot x   cot 2 xdx   n(t  t 2  1 )dt

  t.

[put cotx = t]



= e cosx (x + cosec x) + C



2t

15.

2

2 t  1 dt t  t2  1

t 2



t2  1 ) + t2  1 cos 2x 

cos 2x

cos 2x  x  cot x  cot x n(e(cos x  cos 2x ))  C sin x e x (2  x 2 )

 (1  x)

)

dx  

x dx [(x  5)(2  x)]3 / 2

(5 cos 2   2 sin 2  )( 3 sin 2  ) d [( 3 sin 2  )( 3 cos 2  )]3 / 2



(5 cos 2   2 sin 2  )( 3 sin 2  ) d 27(sin  cos  ) 3

dt

=–n(sinx)cotx–cotx–x – cotxn  cot x  sin x   sin x  

9.

2 3/2



t 1

= – n(sinx)cotx–cotx–x–t n(t +

=

x

 (7x  10  x

1  x2

dx  

e x (1  1  x 2 ) (1  x) 1  x 2

  2   1 1x  dx   ex   1  x   (1  x) 1  x 2        f(x) f '( x )  

dx

  dt 

 C 

(put x = 5 cos 2  + 2 sin 2 )

 n(sin x) cot x  cot x  x  tn(t  t 2  1 ) 

 1 1 t 1   2 3/2  cos t  2 2 (1  t ) 1t 1t 

 1  e t  cos 1 t  1  t2 

  n(sin x ) cot x  cot x  x  n t  t 2  1  t  

1

t



6 (5 cos 2   2 sin 2  ) d 27  (sin  cos  )2



2 (5 cos ec 2   2 sec 2  )d  9 



2 5 cot   2 tan   9 



2 ( 5 cot   2 tan ) d 9 



10 2  x 4  9 x 5 9

x 5 C 2x

EXERCISE - 05 [A] 8.

5 tan x

JEE-[MAIN] : PREVIOUS YEAR QUESTIONS 9.

5 sin x

 tan x  2 dx   sin x  2 cos x dx

I

d 5sinx = A(sinx – 2cosx) + B (sinx – 2cosx) dx = A(sinx – 2cosx) + B (cosx + 2sinx)



3x 2 dx = dt

1 ƒ(t)tdt 3

1 t ƒ(t)dt   (1. ƒ(t)dt)dt   3 

A  2B  5   A = 1, B = 2 2 A  B  0 



5 sin x  sin x  2 cos x dx

Now

put x 3 = t

1 3 x  (x 3 )    (x 3 )x 2 dx 3

 (sin x  2 cos x) 2(cos x  2 sin x)   =   dx sin x  2 cos x   sin x  2 cos x = x + 2log|sinx – 2cosx| + K a = 2

EXERCISE - 05 [B] 1.

 sin

1

JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS

  2x  2   dx 2  4 x  8 x  13 

  x 1   sin 1   2 13  x  2x  4 

 2x  2  3 = (x + 1) tan–1    log(4x2 + 8x + 13)+C  3  4

   dx   

 x 1   sin 1   (x  1) 2  (3 / 2)2

2.

  (x

3m

x

2m

  dx 

 2x 3 m  3x 2 m  6x m   x )  xm   m

 x3m  x2m  x m   x 

3 2 Put x + 1= 3/2 tan, dx  sec d  2

 3    2 tan   1   =  sin   9 9 2  4 tan   4 

I   (x 3 m  x 2 m  x m )(2x 2 m  3x m  6)1 / m dx

 3  sec 2  d  2  

 3m 2m m 1/ m  (2x  3x  6 x ) dx 

Put 2x 3m + 3x 2m + 6x m = y

I 

1 1 y (1 / m ) 1 1/ m y dy  C 6m  6m  mm1  m 1

1 ym  C 6 (m  1)

3 3  sec 2 d    tan    tan d  2 2

3 [ tan   log| sec | ]  C 2 I

 3 2 4  1  2 2  (x  1) tan  (x  1)   log 1  (x  1)   C 2  3 9 3  

 2x  2  3 2 =(x + 1)tan–1    log (9 + 4x + 8x + 4)  3  4

+

3 log 9 + C 4

dx

  (x 3 m 1  x 2 m 1  x m 1 )(2x 3 m  3x 2 m  6 x m )1 / m dx

3  sin  cos   2 sin 1  =   sec  d  2  cos  1 

=

1/m

1 (2x 3 m  3x 2 m  6 x m )  6 m 1 4.

m 1 m

C

Here f f(x) 

f(x) x  [1  f(x) n ]1 / n (1  2x n )1 / n

and f f f(x) 

x (1  3x n )1 / n

 g(x) = (fofo......of)(x) = n 2 Let I   x g(x)dx  

1  2 n

x (1  nx n )1 / n

x n 1 dx (1  nx n )1 / n

n 2 x n 1 1  (1  nx n )1 / n dx  n 2

d (1  nx n ) dx  (1  nx n )1 / n dx

J – I =

Let e x = t =

t2  1  t 4  t2  1 dt =

1 = n 2



sec 2 x dx (sec x  tan x) 9 / 2

sec x(sec x  tan x) sec x dx (sec x  tan x)11 / 2

Put secx + tanx = t  (secx tanx + sec 2x) dx = dt

 secx – tanx =

  e x ex    e 4 x  e 2 x  1 e 4 x  e 2 x  1  dx

e3x  e x dx = =  4x e  e2 x  1

Let I  

Also  sec 2 x – tan 2 x = 1

1

1 1  (1  nx n ) n  K n(n  1)

6.

7.



e x (e 2 x  1)  e 4 x  e 2 x  1 dx

e x dx = dt

1  1 / t2  (t  1 / t)2  1 dt

1 1 x  2x  t + C = 1 n  e  e  1  + C 2 x x 1   2 e  e  1 t  1 t



1

1





1 t

secx =  t   2 t 1  1   dt  1 1 9 / 2 t  t 13 / 2 dt I    11 / 2  . t 2 2 t







1  2t 7 / 2 2t 11 / 2    K 2 7 11 

t

1  1

t2 

=  t11 / 2  11  7   K   =



1 11 / 2

 sec x  tan x 

2 1 1    sec x  tan x    K 11 7 