Test: Integration By Parts


10 Questions MCQ Test Mathematics (Maths) Class 12 | Test: Integration By Parts


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QUESTION: 1

Evaluate: 

Solution:
QUESTION: 2

Evaluate: 

Solution:
QUESTION: 3

Solution:

 ∫ex[tan x + log sec x] dx
=∫ex . tan x dx + ∫ex . log sec x dx
= ex .∫ tan x dx − ∫[ddx(ex)×∫ tan x dx] dx + ∫ex . log sec x dx
= ex . log sec x − ∫ex . log sec x dx + ∫ex . log sec x dx + C
= ex . log sec x + C

QUESTION: 4

Evaluate : 

Solution:
QUESTION: 5

Solution:

 q = √x, dq = dx/2√x
⇒ dx = 2q dq
so the integral is 2∫qcosqdq
integration by parts using form 
∫uv' = uv − ∫u'v
here u = q, u'= 1 and v'= cosq, v=sinq
so we have 2(qsinq −∫sinqdq)
= 2(qsinq + cosq + C)
= 2(√xsin√x + cos√x + C)

QUESTION: 6

Evaluate:

Solution:

Let  x=tanθ , so that  θ=tan−1x ,  dx=sec2θdθ 
Then the given integral is equivalent to
∫tan−1(2x/(1−x2)dx
=∫tan−1 (2tanθ/(1−tan2θ))⋅sec2θdθ 
=∫tan−1tan2θ⋅sec2θdθ 
=2×∫θsec2θdθ 
(integrate by parts)
=2θ⋅∫sec2θdθ −2⋅∫1⋅(∫sec2θdθ)dθ 
=2θtanθ−2⋅∫tanθdθ 
=2θtanθ−2loge secθ+c 
=2xtan−1x−2loge√1+x2+c 
=2xtan−1x−loge(1+x2)+c 

QUESTION: 7

The integration of the function ex.cos3x is:

Solution:

Let I = ∫ex . cos 3x dx
⇒ I = cos 3x × ∫ex dx − ∫[d/dx(cos 3x) × ∫ex dx]dx
⇒ I = ex cos 3x − ∫(− 3 sin 3x . ex)dx
⇒ I = ex cos 3x + 3∫sin 3x . ex dx
⇒ I = ex cos 3x + 3[sin 3x × ∫ex dx − ∫{ddx(sin 3x) × ∫ex dx}dx]
⇒ I = ex cos 3x + 3[sin 3x . ex − ∫3 cos 3x . ex dx]
⇒ I = ex cos 3x + 3 ex . sin 3x − 9∫ex . cos 3x dx
⇒ I =  ex cos 3x + 3 ex . sin 3x − 9I
⇒ 10I =  ex cos 3x + 3 ex . sin 3x
⇒ I = 1/10[ex cos 3x + 3 ex . sin 3x] + C

QUESTION: 8

If f(x) and g(x) are two functions of x, then the integration of product of f(x) and g(x) is:

Solution:
QUESTION: 9

Evaluate: 

Solution:

I=∫sin(logx)×1dx
= sin(logx) × x−∫cos(logx) × (1/x)×xdx
= xsin(logx)−∫cos(logx) × 1dx
= xsin(logx)−[cos(logx) × x−∫sin(logx) × (1/x) × xdx]
∴ I=xsin(logx)−cos(logx) × x−∫sin(logx)dx
2I=x[sin(logx)−cos(logx)]
∴ I=(x/2)[sin(logx)−cos(logx)]

QUESTION: 10

Evaluate:

Solution:

(tan-1 x)x dx
= tan-1 ∫x dx - ∫d(tan-1 x)/dx ∫x.dx)dx
= tan-1 x . (x2)/2 - ∫1/(1+x2). (x2)/2 dx
=(x2)/2 tan-1 x - 1/2∫(x2)/(x2 + 1) dx
= (x2)/2 tan-1 x - 1/2∫(x2 + 1 - 1)/(x2 + 1) dx
= (x2)/2 tan-1 x - ½[∫(x2 + 1)/(x2 + 1)dx - dx/(x2 + 1)]
= (x2)/2 tan-1 x - ½[dx - ∫dx/(x2 + 1)]
By applying formula, ∫dx/(a2 + x2) = 1/a tan-1x/a + c, we get
= (x2)/2 tan-1 x - x/2 + ½ * 1/1 tan-1 (x) + c
= (x2)/2 tan-1 x - x/2 + ½ tan-1 (x) + c

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