Mathematics_Today_-_October_2016

2. (c) : Let I = cos
   cos

21

21

x

x

− dx

∫ +

⇒ =− −

• 
  

I ∫∫x =−

x

dx x

x

(cos) dx

(cos)

sin

cos

12

12

2

2

2

2

⇒ Ixdx xdx=−∫tan^22 =−−∫(sec 1 )

⇒ I = x – tanx + c

3. (b) : ()
   () ()

x

xx

dx xx

xx

+ dx

+

= ++

∫∫+

1

1

12

1

2

2

2

2

= +

+

+

∫∫+

x

xx

dx x

xx

dx

2

22

1

1

2

() () 1

=+

+

∫dx ∫ =+− +

x

dx

x

(^2) exxc
1
2 log^2 tan^1

4. (c) : Put xe + ex = t ⇒ e(xe – 1 + ex – 1)dx = dt

Now, xe

xe

dx

e

dt

te

t

e

xe c

ex

ex

−−+ ex

+

∫∫== = ++

(^11) 11 1
log log( )

5. (b) : LetI x
   xx

= dx

∫ +

sin

sin cos

2

44

=

+

=

∫ ∫ +

22

(^441)
2
4
sin cos
sin cos
tan sec
tan
xx
xx
dx xx
x
dx
Put tan^2 x = t ⇒ 2tanx sec^2 xdx = dt
∴ I = dt
t
tc x c
1 2
112

• ∫ =+=tan−−tan (tan )+

6. (a) : Put a^2 + b^2 sin^2 x = t ⇒ b^2 sin2xdx = dt, then
   sin
   sin
   222211 x 2 2 log
   ab x

dx

b

dt

t b

tc

+

∫∫==+

=++^12222

b

log(absin xc)

7. (c) :

cos

(cos sin )

2

2

x

xx

dx

∫ +

= − +

∫ +

(cos sin )(cos sin )

(cos sin )

xxxx

xx

2 dx= −

∫ +

cos sin

cos sin

xx

xx

dx

Put t = sinx + cosx ⇒dt =(cosx – sinx)dx, then it
reduces to
1
t
∫ dt=+=logt c log(sinx+ +cos )x c

8. (a) :^1
   11 −^22

=

−

∫∫

−

e dx −

e

e

x dx

x

x

Put e–x = t ⇒ –e–x dx = dt, then it reduces to

−

−

∫^1 =− + − +

1

2 2 1

t

dt log[t t ] c

=− + − =− + −

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥

log[ee−−] log

e

e

e

xx

x

x

x

2 1 112

=−log[ 11 + −eec^2 xx] log++ =xec−log[ 11 + −^2 x]+

9. (c) : sin
   sin sin

sin( )

sin sin

2

53

53

53

x

xx

dx xx

xx

∫∫= − dx

=∫sin cos −cos sin

sin sin

53 53

53

xx xx

xx

dx

=^1 − +

3

3 1

5

log sin xxclog sin 5

10. (b) : dx
    ∫xxlog ⋅log(log )x
    Put log (logx) = z
    ⇒ 11
    logxx

⋅ dx=dz, then it reduces to

dz

z

∫ ==logzxclog[log(log )]+

11. (a) : Put x = sinθ ⇒ dx = cosθdθ, then
    1
    1

1 1

2

12

2

2

+ 2

−

∫∫x =+∫ =+ −

x

dx ( sin θθθ)d ( cos θθ)d

=^3 −−+

2

1

2

θ sinθθ 1 sin (^2) c= (^3) − −−+
2
1
2
sin^12 xx xc 1

12. (b) : Put x = a(sinθ)2/3
    ⇒ dx=^2 a − d
    3

(sin )θθθ^13 / cos

∴

−

=

−

∫∫

−

x

ax

dx

aa

aa

33 d

12 13 13

332

2

3

//(sin ) (sin ) /cos

sin

θθθ

θ

θ

=

−

(^2) ∫ = − ⎜⎛⎝ ⎞⎟⎠ +
(^31)
2
3
32
32 2
1
32
a d
a
x
a
/ c
/
cos /
sin
θθ sin
θ

13. (d) : Put 1 + x^3 = t^2 ⇒ 3 x^2 dx = 2tdt
    ∴ Itdt=^2 ∫ −
    3

()^21

= −

⎛

⎝

⎜

⎞

⎠

⎟+= − += − ++

2

33

2

9

3 2

9

21

t^3233

tc tt() ( )c x xc

14. (a) : LetI
    xx

= dx

∫ − +

1

[( 12 )(^3514 )]/

=

−

+

⎛

⎝⎜

⎞

⎠⎟ +

−

+ = ⇒ + =

∫

1

1

2

2

1

2

3

2

(^342)
2
x
x
x
dx
x
x t x dx dt
/
()
()
Put