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All questions of Chapter 7 - Integrals for JEE Exam
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
a)
b)
c)
d)
|
Yogesh Singla answered |
D is right sinx= -cox and x*n=1/n x*n+1
Integral of sin5x.cos2x is: - a)
- b)
- c)
- d)
Correct answer is 'A'. Can you explain this answer?
Integral of sin5x.cos2x is:
a)
b)
c)
d)
|
Learners Habitat answered |
So Looking at this integral, we have
- 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?
a)
cos(sin-1x) + c
b)
sin-1x + c
c)
sin(cos-1x) + c
d)
x + c
|
Patel Smit answered |
Take sin inverse x is equal to t and 1 upon under root 1 minas x square is equal to dt,then integration of cost is equal to sint plus c put t I equal to sin inverse x and got your answer.
- 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?
a)
, where C is a constantb)
, where C is a constantc)
, where C is a constantd)
, where C is a constant
|
Abha Dhamija 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)
⇒ 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)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Evaluate:
a)
b)
c)
d)
|
Krishna Iyer answered |
∫dx/x(xn + 1)..............(1)
∫dx/x(xn + 1) *(xn - 1)/(xn - 1)
Put xn = t
dt = nx(n-1)dx
dt/n = x(n-1)dx
Put the value of dt/n in eq(1)
= ∫(1/n)dt/t(t+1)
= 1/n ∫dt/(t+1)t
= 1/n{∫dt/t - ∫dt/t+1}
= 1/n {ln t - lnt + 1} + c
= 1/n {ln |t/(t + 1)|} + c
= 1/n {ln |xn/(xn + 1)|} + c
∫dx/x(xn + 1) *(xn - 1)/(xn - 1)
Put xn = t
dt = nx(n-1)dx
dt/n = x(n-1)dx
Put the value of dt/n in eq(1)
= ∫(1/n)dt/t(t+1)
= 1/n ∫dt/(t+1)t
= 1/n{∫dt/t - ∫dt/t+1}
= 1/n {ln t - lnt + 1} + c
= 1/n {ln |t/(t + 1)|} + c
= 1/n {ln |xn/(xn + 1)|} + c
The integration of the function ex.cos3x is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The integration of the function ex.cos3x is:
a)
b)
c)
d)
|
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
⇒ 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
Evaluate as limit of sum 
- a)20/5
- b)15/2
- c)20/3
- d)3/20
Correct answer is option 'C'. Can you explain this answer?
Evaluate as limit of sum
a)
20/5
b)
15/2
c)
20/3
d)
3/20
|
|
Om Desai answered |
∫(0 to 2)(x2 + x + 1)dx
= (0 to 2) [x3/3 + x2/2 + x]½
= [8/3 + 4/2 + 2]
= 40/6
= 20/3
= (0 to 2) [x3/3 + x2/2 + x]½
= [8/3 + 4/2 + 2]
= 40/6
= 20/3
Evaluate: 
- a)1/2
- b)1/4
- c)1
- d)1/8
Correct answer is option 'B'. Can you explain this answer?
Evaluate:
a)
1/2
b)
1/4
c)
1
d)
1/8
|
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
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
- a)-1
- b)zero
- c)1
- d)2
Correct answer is option 'B'. Can you explain this answer?
a)
-1
b)
zero
c)
1
d)
2
|
Naina Sharma 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
= [[(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
is: 
The value of 
is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The value of is:
is:a)
b)
c)
d)
|
Samyak Jain answered |
1/2(cos π/2 + 1 -[cos(π/2-π/4)+1])=1/2+π/4
The integral of 
is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The integral of is:
is:a)
b)
c)
d)
|
|
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
= ∫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?
Evaluate:
a)
b)
c)
d)
|
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
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
Evaluate: 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Evaluate:
a)
b)
c)
d)
|
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)]
= 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)]
If 
is- a)2/3
- b)4/5
- c)1
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
If is
isa)
2/3
b)
4/5
c)
1
d)
None of these
|
Knowledge Hub answered |
In the question, it should be f’(2) instead of f”(2)
Explanation:- f(x) = ∫(0 to x) log(1+x2)
f’(x) = 2xdx/(1+x2)
f’(2) = 2(2)/(1+(2)2)
= 4/5
Explanation:- f(x) = ∫(0 to x) log(1+x2)
f’(x) = 2xdx/(1+x2)
f’(2) = 2(2)/(1+(2)2)
= 4/5
then the value of k is:
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
a)
b)
c)
d)
|
|
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
∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx
is:
is:
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?
The value of the integral is:
is:a)
2e – 1
b)
2e + 1
c)
2e
d)
2(e – 1)
|
Aryan Khanna 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)(x + b) cos(a – b) – sin(a – b).log | sin(x + b) | + C
- b)(x + b) cos(b – a) – sin(b – a).log | sin(x + b) | + C
- c)(x + b) cos(b – a) + sin(b – a).log | sin(x + b) | + C
- d)(x + b) cos(a – b) + sin(a – b).log | sin(x + b) | + C
Correct answer is option 'D'. Can you explain this answer?
Evaluate:
a)
(x + b) cos(a – b) – sin(a – b).log | sin(x + b) | + C
b)
(x + b) cos(b – a) – sin(b – a).log | sin(x + b) | + C
c)
(x + b) cos(b – a) + sin(b – a).log | sin(x + b) | + C
d)
(x + b) cos(a – b) + sin(a – b).log | sin(x + b) | + C
|
Tanuja Kapoor answered |
Create ((X+B) + (A-B)) in numerator and then apply Sin(a+b) formula then you will be able to solve it.
The value of 
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The value of
a)
b)
c)
d)
|
Shubham Rajput answered |
I have got correct answer as d please confirm me anyone
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?
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
|
|
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
= ∫(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
Evaluate: 
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Evaluate:
a)
b)
c)
d)
|
|
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
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
- 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?
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
|
|
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
⇒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 integral of tan4x is:- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The integral of tan4x is:
a)
b)
c)
d)
|
|
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,
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
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
∫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
∫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
∫tan2xdx = tanx − x + C
Combining the two results, we have:
∫tan4xdx=∫sec2xtan2xdx−∫tan2xdx
∫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.
- a)if f(2a – x) = – f(x)
- b)if f(2a – x) = f(x)
- c)if f(- x) = f(x)
- d)if f(- x) = – f(x)
Correct answer is option 'D'. Can you explain this answer?
a)
if f(2a – x) = – f(x)
b)
if f(2a – x) = f(x)
c)
if f(- x) = f(x)
d)
if f(- x) = – f(x)
|
|
Om Desai answered |
∫(-a to a)f(x)dx
= ∫(0 to a) [f(x) + f(-x)] if f(x) is an odd function
⇒ f(-x) = -f(x)
= ∫(0 to a) [f(x) + f(-x)] if f(x) is an odd function
⇒ f(-x) = -f(x)
is:
- 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?
a)
g (x) h (s) = constant
b)
g (x) = h (x).
c)
g (x) - h (x) = constant
d)
h (x) + g (x) = constant
|
Swati Verma answered |
Since g(x) and h(x) are integrals of the same function , therefore ; g(x) – h(x) is constant.
Find the distance travelled by a car moving with acceleration given by a(t)=Sin(t), if it moves from t = 0 sec to t = π/2 sec, if velocity of a car at t=0sec is 10 km/hr.- a)10.19 km
- b)19.13 km
- c)15.13 km
- d)13.13 km
Correct answer is option 'D'. Can you explain this answer?
Find the distance travelled by a car moving with acceleration given by a(t)=Sin(t), if it moves from t = 0 sec to t = π/2 sec, if velocity of a car at t=0sec is 10 km/hr.
a)
10.19 km
b)
19.13 km
c)
15.13 km
d)
13.13 km
|
Rohit Yadav answered |
Add constant automatically
We know that,
We know that,
Find ∫6x(x2+6)dx.- a)5x3 + 9x² + c.
- b)10x5 + 9x3 + c.
- c)1.5x⁴ + 18x² + c.
- d)3.5x4 + 11x3 + c.
Correct answer is option 'C'. Can you explain this answer?
Find ∫6x(x2+6)dx.
a)
5x3 + 9x² + c.
b)
10x5 + 9x3 + c.
c)
1.5x⁴ + 18x² + c.
d)
3.5x4 + 11x3 + c.
|
|
Maitri Roy answered |
The word "find" is a verb that means to discover something or someone through searching, examining, or investigating. It can also mean to come across or encounter something unexpectedly or accidentally. Additionally, "find" can refer to forming an opinion or reaching a conclusion about something or someone based on examination or investigation.
Read the following text and answer the following questions on the basis of the same:
Q. 
- a)1
- b)0
- c)-1
- d)π
Correct answer is option 'B'. Can you explain this answer?
Read the following text and answer the following questions on the basis of the same:
Q.
a)
1
b)
0
c)
-1
d)
π
|
|
Anjali Sharma answered |
since sin3 x is an odd function.
- a)log 2
- b)0
- c)
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
a)
log 2
b)
0
c)
d)
none of these
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Shilpa Saha answered |
∫ log(1-x) dx - ∫ log x dx = 0
Chapter doubts & questions for Chapter 7 - Integrals - Mathematics CUET UG Mock Test Series 2025 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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