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All questions of Integration for EmSAT Achieve Exam

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  • a)
    cos(sin-1x) + c
  • b)
    sin-1x + c
  • c)
    sin(cos-1x) + c
  • d)
    x + c
Correct answer is option 'D'. Can you explain this answer?

Divey Sethi answered
cos(sin-1x)/(1-x2)½……………….(1)
t = sin-1 x
dt = dx/(1-x2)½
Put the value of dt in eq(1)
= ∫cost dt
= sint + c
= sin(sin-1 x) + c
⇒ x + c

Evaluate: 
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Leelu Bhai answered
I = ∫√(x² + 5x)dx= ∫√(x² + 5x + 25/4 - 25/4)= ∫√{(x + 5/2)² - (5/2)²}={1/2(x+5/2)(√x² + 5x)} - {25/8 log{(x + 5/2)+√x²+ 5x}}= {(2x + 5)/4 (√x² + 5x)} - {25/8 log{(x + 5/2)+√x²+ 5x}}Thus, option D is correct...

Evaluate:  
  • a)
  • b)
    1/√3 arc tan[(x-2)/√5] + C
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Deepak Kapoor answered
  
Let's apply the integral substitution,
substitute 
Now use the standard integral :
substitute back u=(x-2) and add a constant C to the solution,
 

  • a)
    , where C is a constant
  • b)
    , where C is a constant
  • c)
    , where C is a constant
  • d)
    , where C is a constant
Correct answer is option 'C'. Can you explain this answer?

Neha Sharma answered
 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)

Evaluate: 
  • a)
    1/2
  • b)
    1/4
  • c)
    1
  • d)
    1/8
Correct answer is option 'B'. Can you explain this answer?

Sumair Sadiq answered
This is maths questions I can explain it but you know it is not possible here because this app is not allow to take photo but try it ok let tan inverce 4x =t diff both side wrt x 4x³/1+x⁴ Ka square
x cube / 1+ x8 =dt/4 I = £ 0 se pie by 2 (because when x= 0 t = pie by 2and x = infinity then t = 0 )
I = 1/4 £ 0 se pie by 2 sin t l = 1/4 (- cos t limit 0 se pie by 2 )l = 1/4 ( - cos pie by 2 + cos 0) l = 1/4 ( 0+ 1) l= 1/4 × 1l= 1/4
use my WhatsApp number for further questions but only for study 7060398771

Evaluate: 
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Preeti Iyer answered
 (x)½ (a - x)½ dx
=  ∫(ax - x2)½ dx
=  ∫{-(x2 - ax)½} dx
=  ∫{-(x2 - ax + a2/4 - a2/4)½} dx
=  ∫{-(x - a/2)2 - a2/4} dx
=  ∫{(a/2)2 - (x - a/2)2} dx
=  ½(x - a/2) {(a/2)2 - (x - a/2)2} + (a2/4) (½ sin-1(x - a)/a2)
= {(2x - a)/4 (ax - x2)½} + {a2/8 sin-1(2x - a)/a} + c

The integration of the function ex.cos3x is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Om Desai answered
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

  • a)
    -1
  • b)
    zero
  • c)
    1
  • d)
    2
Correct answer is option 'B'. Can you explain this answer?

Praveen Kumar answered
∫(0 to 4)(x)1/2 - x2 dx
= [[(x)3/2]/(3/2) - x2](0 to 4)
= [[2x3/2]/3 - x2](0 to 4)
= [[2(0)3/2]/3 - (0)2]] -  [[2(4)3/2]/3 - (4)2]]
= 0-0
= 0

The value of the integral is:
  • a)
    2e – 1
  • b)
    2e + 1
  • c)
    2e
  • d)
    2(e – 1)
Correct answer is option 'D'. Can you explain this answer?

Infinity Academy answered
Correct Answer : d
Explanation :  ∫(-1 to 1) e|x| dx
∫(-1 to 0) e|x|dx + ∫(0 to 1) e|x|dx
 e1 -1 + e1 - 1
=> 2(e - 1)

Evaluate: 
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Vikas Kapoor answered
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)]

Integrate 
  • a)
    3x – 4 log |sec x| + tan x + C
  • b)
    3x + 4 log |sec x| + tan x
  • c)
    3x + 4 log |sec X| + tan x + C
  • d)
    3x + 4 log |sec x| – tan x + C
Correct answer is option 'C'. Can you explain this answer?

Vikas Kapoor answered
∫(2+tan x)2dx
= ∫(4 + tan2 x + 4tan x)dx
= ∫4 dx + ∫tan2 x dx + 4∫tan x dx
= 4x + ∫(sec2 x - 1)dx + 4(log|sec x|)
= 4x + tanx - x + 4(log|sec x|)
3x + tanx + 4(log|sec x|) + c

  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Vikas Kapoor answered
Option d is correct, because it is the property of definite integral
 ∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx

  • a)
    π
  • b)
    π/2
  • c)
    2π
  • d)
    π/4
Correct answer is option 'B'. Can you explain this answer?

Tarun Kaushik answered
For sin2(X), we will use the cos double angle formula:
cos(2X) = 1 - 2sin2(X)
The above formula can be rearranged to make sin2(X) the subject:
sin2(X) = 1/2(1 - cos(2X))
You can now rewrite the integration: 
∫sin2(X)dX = ∫1/2(1 - cos(2X))dX
Because 1/2 is a constant, we can remove it from the integration to make the calculation simpler. We are now integrating:
1/2 x ∫(1 - cos(2X)) dX 
= 1/2 x (X - 1/2sin(2X)) + C]-pi/4 to pi/4
∫sin2(X) dX = [1/2X - 1/4sin(2X)]-pi/4 to pi/4 + C
½[-pi/2] - 1/4sin(2(-pi/4)] - ½[pi/2] - 1/4sin(2(pi/4)] 
= π/2

Evaluate: 
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Om Desai answered
sin2x = 1 - cos2x
∫sinx(sin2x - 3cos2x + 15)dx
Put cos2x = t
 ∫sinx(1 - cos2x - 3cos2x + 15)dx
=  ∫sinx (16 - 4cos2x)dx
Put t = cosx, differentiate with respect to x, we get 
dt/dx = -sinx
= -  ∫ [(16 - 4t2)]1/2dt
= -2 ∫ [(2)2 - (t)2]½
= -2{[(2)2 - (t)2]½ + 2sin-1(t/2)} + c
= - cosx {[4 - (cos)2x]½ - 4sin-1(cosx/2)} + c

  • a)
    log(sin x + cos x) +C
  • b)
    sin 2x + cos 2x + C
  • c)
    log(sin x - cos x) +C
  • d)
    sin 2x - cos 2x + C
Correct answer is option 'A'. Can you explain this answer?

Om Desai answered
 I = ∫cos2x/(sinx+cosx)2dx
⇒I = ∫cos2x−sin2x(sinx+cosx)2dx
⇒I = ∫[(cosx+sinx)(cosx−sinx)]/(sinx+cosx)2dx
⇒I = ∫(cosx−sinx)/(sinx+cosx)dx
Let sinx+cosx = t 
(cosx−sinx)dx = dt
Then, I = ∫dt/t
I = log|t|+c
I = log|sinx + cosx| + c

The value of  is:
  • a)
    10
  • b)
    17/2
  • c)
    7/2
  • d)
    5
Correct answer is option 'A'. Can you explain this answer?

Sushil Kumar answered
∫(-3 to 3) (x+1)dx
=  ∫(-3 to -1) (x+1)dx +  ∫(-1 to 3) (x+1) dx 
= [x2 + x](-3 to -1) + [x2 + x](-1 to 3)
= [½ - 1 - (9/2 - 3)] + [9/2 + 3 - (½ - 1)]
= -[-4 + 2] + [4 + 4]
= -[-2] + [8]
= 10

If   then the value of k is:
  • a)
    7/8
  • b)
    5/8
  • c)
    1/2
  • d)
    3/2
Correct answer is option 'C'. Can you explain this answer?

Sushil Kumar answered
Let I=∫(0 to k) 1/[1 + 4x2]dx = π8
Now, ∫(0 to k) 1/[4(1/4 + x2)]dx
= 2/4[tan−1 2x]0 to k
= 1/2tan-1 2k − 0 = π/8
1/2tan−1 2k = π8
⇒ tan−1 2k = π/4
⇒ 2k = 1
∴ k = 1/2

Evaluate:  
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Om Desai answered
 ∫sin2(2x+1) dx
Put t = 2x+1
dt = 2dx
dx= dt/2
= 1/2∫sin2 t dt
=1/2∫(1-cos2t)/2 dt
= 1/4∫(1-cos2t) dt
= ¼[(t - (sin2t)/2]dt
= t/4 - sin2t/8 + c
= (2x+1)/4 - ⅛(sin(4x+2)) + c
= x/2 - 1/8sin(4x+2) + ¼ + c
As ¼ is also a constant, so eq is = x/2 - 1/8sin(4x+2) + c

The integral of tan4x is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Aryan Khanna answered
Begin by rewriting  ∫tan4xdx as ∫tan2xtan2xdx.      
Now we can apply the Pythagorean Identity,  tan2x+1=sec2x,                                         or tan2x=sec2x−1

∫tan2x tan2x dx = ∫(sec2x−1)tan2xdx

Distributing the tan2x:
             = ∫sec2xtan2x − tan2xdx
Applying the sum rule:
                = ∫sec2xtan2xdx − ∫tan2xdx
We'll evaluate these integrals one by one.
First Integral 
This one is solved using a 
Let u = tanx


Applying the substitution,
Because u = tanx,
Second Integral
Since we don't really know what  ∫tan2xdx is by just looking at it, try applying the tan2x = sec2x−1 
identity again:
∫tan2xdx = ∫(sec2x−1)dx
Using the sum rule, the integral boils down to:
∫sec2xdx − ∫1dx
The first of these,  ∫sec2xdx, is just tanx + C.
The second one, the so-called "perfect integral", is simply x+C.
Putting it all together, we can say:
∫tan2xdx = tanx + C − x + C
And because C+C is just another arbitrary constant, we can combine it into a general constant C:
∫tan2xdx = tanx − x + C
Combining the two results, we have:
∫tan4xdx=∫sec2xtan2xdx−∫tan2xdx
=(tan3x/3 + C) − (tanx − x + C)
=tan3x/3 − tanx + x + C
Again, because C+C is a constant, we can join them into one C.

The integral of   is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Nandini Iyer answered
 ∫dx/x3(x-2 -4).............(1)
=  ∫x-3 dx/(x-2 - 4)​
Let t = (x-2 - 4)
dt = -2x-3 dx
x-3 = -dt/2
Put the value of x-3 in eq(1)
= -½  ∫dt/t
= -½ log t + c
= -½ log(x-2 - 4) + c
= -½ log(1-4x2)/x2 + c
= ½ log(x2/(1 - 4x2)) + c

Evaluate:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'D'. Can you explain this answer?

Hansa Sharma answered
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 

  • a)
    g (x) h (s) = constant
  • b)
    g (x) = h (x).
  • c)
    g (x) - h (x) = constant
  • d)
    h (x) + g (x) = constant
Correct answer is option 'C'. Can you explain this answer?

Seblewongel Girma answered
Integration of a function can have many possibility by adding variable C.
for instance :
take integral of X^2=X^3/3 +C
integral of X^2 can be X^3/3 +1, X^3/3 +2.....
their d/CE is a constant.

Chapter doubts & questions for Integration - Mathematics for EmSAT Achieve 2025 is part of EmSAT Achieve exam preparation. The chapters have been prepared according to the EmSAT Achieve exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for EmSAT Achieve 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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