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∫ \(\frac{1+tan\,x\,.\,tan(x+\theta)}{[tan(x+\theta)-tan\,x]}\) (tan(x+θ) - tan x)dx

= ∫ \(\frac1{tan\,\theta}\) [tan(x+θ) - tan x] . dx

= \(\frac1{tan\,\theta}\) ∫ tan(x+θ)dx - ∫ tan x . dx

= \(\frac1{tan\,\theta}\) [-log cos(x+θ)1] + [log |cos x|] + c

= \(\frac1{tan\,\theta}\) log |\(\frac{cos\,x}{cos(x+\theta)}\)| + c

= cot θ log |\(\frac{cos\,x}{cos(x+\theta)}\)| + c.

tan⁷x/7  +  C

To help the students in their preparations, Class 12 Maths MCQ Questions of Integrals with Answers were created by subject specialists as per exam pattern and syllabus. The ideas of integrals are given in an exhaustive and straightforward manner. These Important MCQ Questions are exceptionally straightforward and can without much of a stretch help students in learning the interaction of problem-solving.

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Practice MCQ Question for Class 12 Maths chapter-wise

 1. ∫ 1+sinx/1+cosx . e^x dx is equal to

(a) e^x tan(x/2)+k
(b) e^x tanx+k
(c) 1/2e^x tan(x/2)+k
(d) e^x sec²(x/2)+k

2. \(\int\frac{2^x}{\sqrt{1-4^x}}dx\) = K sin-1(2x)+C, then K is equal to

(a) ln2
(b) 1/2 ln2
(c) 1/2
(d) 1/ln2

3. If \(\int\frac{2^{1/x}}{x^2}dx=\) k.21/x+C, then k is equal to

(a) -1/log_e2
(b) -log_e2
(c) -1
(d) 1/2

4. \(\int\frac{1}{sin^2xcos^2x}dx\) is equal to

(a) sin² x – cos² x + C
(b) -1
(c) tan x + cot x + C
(d) tan x – cot x + C

5. \(\int\frac{cos2x-cos2\theta}{cosx-cos\theta}dx\) is equal to

(a) 2(sin x + x cos θ) + C
(b) 2(sin x – x cos θ) + C
(c) 2(sin x + 2x cos θ) + C
(d) 2(sin x – 2x cos θ) + C

6. ∫cot²x dx equals to

(a) cot x – x + C
(b) cot x + x + C
(c) -cot x + x + C
(d) -cot x – x + C

7. \(\int\frac{sinx+cosx}{\sqrt{1+sin2x}}dx,\frac{3\pi}{4}<x<\frac{7\pi}{4}\) is equal to

(a) log |sin x + cos x|
(b) x
(c) log |x|
(d) -x

8. If ∫ sec²(7 – 4x)dx = a tan (7 – 4x) + C, then value of a is

(a) 7
(b) -4
(c) 3
(d) −1/4

9. The value of X for which

\(\int\frac{4x^3+\lambda4^x}{4^x+x^4}dx=log|4^x+x^4|is\)

(a) 1
(b) log_e4
(c) loe₄ e
(d) 4

10. If \(\int\frac{1}{\sqrt{4-9x^2}}dx=\frac{1}{3}sin^{-1}(ax)+C,\) then value of a is

(a) 2
(b) 4
(c) 3/2
(d) 2/3

11. \(\int\frac{10x^9+10^xlog_e10}{x^{10}+10^x}dx\) equals

(a) 10^x -x¹⁰ + c
(b) 10^x + x¹⁰ + c
(c) (10^x – x¹⁰)^-1 + c
(d) log (10^x + x¹⁰) + c.

12. \(\int\frac{e^x(1+x)}{cos^2(xe^2)}dx \) is equals to

(a) -cot (xe^x) + c
(b) tan (xe^x) + c
(c) tan (e^x) + c
(d) cot (e^x) + c

13. If \(\int\frac{1}{(x^2+4)(x^2+9)}dx=A\,tan^{-1}\frac{x}{2}+B\,tan^{-1}(\frac{x}{3})\) +C then A-B =

(a) 1/6
(b) 1/30
(c) -1/30
(d) -1/6

14. The value of \(\int\frac{e^x(x^2tan^{-1}x+tan^{-1}x+1)}{x^2+1}\)  dx is equal to

(a) e^xtan^-1x+C
(b) tan^-1(e^x)+C
(c) tan^-1(x^e)+C
(d) e^{tan^-1}x+C

15. If ∫ dx/[(x+2)(x²+1)] =  a log |1 + x²| + b tan^–1x + (1/5) log |x + 2| + C, then 

(a) a = -1/10, b = -2/5 
(b) a = 1/10, b = -2/5 
(c) a = -1/10, b = 2/5 
(d) a = 1/10, b = 2/5

16. \(\int x^2e^{x^3}\) dx equals

(a) 1/3 e^{x^3}+C
(b) 1/3 e^{x^4}+C
(c) 1/2 e^{x^3}+C
(d) 1/2 e^{x^2}+C

17. ∫e^x sec x (1 + tan x) dx equals

(a) e^x cos x + c
(b) e^x sec x + c
(c) e^x sin x + c
(d) e^x tan x + c.

18. ∫tan^-1 √x dx is equal to

(a) (x + 1)tan^-1 √x – √x + c
(b) x tan^-1 √x – √x + c
(c) √x – x tan^-1 √x + c
(d) tan^-1x – (x + 1) tan^-1 √x + c

19. ∫e^x(cosx−sinx)dx is equal to

(a) e^x cosx + C
(b) e^x sinx + C
(c) -e^x cosx + C
(d) -e^x sinx + C

20. If a is such that \(\int_0^axdx\) ≤ a + 4, then

(a) 0 ≤ a ≤ 4
(b) -2 ≤ a ≤ 0
(c) a ≤ -2 or a ≤ 4
(d) -2 ≤ a ≤ 4

Answer: 

1. Answer: (a) e^x tan(x/2)+k

Explanation: \(\int\frac{1+sin\,x}{1+cos\,x}.e^xdx\)

Use \(cos\,x=\frac{1-tan^2(x/2)}{1+tan^2(x/2)},sin\,x=\frac{2tan(x/2)}{1+tan^2(x/2)}\)

The integral becomes

\(=\int\frac{1+tan^2(x/2)+2\,tan(x/2)}{1+tan^2(x/2)+1-tan^2(x/2)}.e^xdx\)

\(=\int\frac{sec^2(x/2)+2\,tan(x/2)}2.e^xdx\)

\(=\int\left[\frac12sec^2(x/2)+tan(x/2)\right]e^xdx\)

If we take tan (x/2) = f(x), the equation takes the form

\(\int e^x\) [f(x) + f'(x)]dx

\(=e^x\) f(x) + k

\(=e^x\) tan (x/2) + k

2. Answer: (d) 1/ln2

Explanation: \(I=\int\frac{2^x}{\sqrt{1-4^x}}dx\)

Let \(2^x\) = z

\(\Rightarrow\) \(2^x\) log 2dx = dz

\(\Rightarrow\) \(I=\int\cfrac{\frac1{log\,2}}{\sqrt{1-z^2}}dz\)

\(=\frac1{log\,2}.sin^{-1}z+c\)

\(=\frac1{log\,2}.sin^{-1}(e^x)+c\)

\(=k\,sin^{-1}(2^x)+c\)

then k \(=\frac1{log\,2}.\)

3. Answer: (a) -1/log_e2

Explanation: \(I=\int\frac{2^{\frac1x}}{x^2}dx=k.2^{\frac1x}+C\;\;...(1)\)

Let \(\frac1x\) = t

\(\Rightarrow\) \(\frac{-1}{x^2}dx=dt\)

\(\Rightarrow\) \(\frac{dx}{x^2}\) = -dt

Put \(\frac1x\) = t and \(\frac{dx}{x^2}\) = -dt  ....in LHS of equation (1) we get

\(I=-\int2^t.dt\)

\(=-\frac{2^t}{In\,2}+C\)

\(=-\frac{2^{\frac1x}}{In\,2}+C\;\;...(2)\)

Comparing of RHS of equation (1) with equation (2) we get

\(k=-\frac1{In\,2}\)

\(\therefore\) \(k=-\frac1{log_e\,2}\)

4. Answer: (d) tan x – cot x + C

Explanation: \(\int\frac1{sin^2x\,cos^2x}dx\)

\(=\int\frac{sin^2x+cos^2x}{sin^2x\,cos^2x}dx\)

\(=\int(sec^2x+cosec^2x)dx\)

= tan x - cot x + C

5. Answer: (a) 2(sin x + x cos θ) + C

Explanation: as \(\int\frac{2(cos^2x-cos^2\theta)}{cos\,x-cos\,\theta}dx,\)

using cos 2x = 2 \(cos^2x\) - 1

= 2 \(\int(cos\,x+cos\,\theta)dx\)

= 2 sin x + 2x . cos \(\theta\) + C

6. Answer: (d) -cot x – x + C

Explanation: ∫ (cosec²x -1)dx = -cot x – x + C

7. Answer: (d) -x

Explanation: as \(\int\frac{sin\,x+cos\,x}{|sin\,x+cos\,x|}dx,\)

\(\Rightarrow\) \(-\int\) 1.dx = -x + C

{as sin x + cos x < 0 for \(\frac{3\pi}{4}<x<\frac{7\pi}{4}\}\)

8. Answer: (d) −1/4

Explanation: \(\int sec^2(7-4x)dx=\frac{tan(7-4x)}{-4}+C\)

\(=-\frac14\) tan (7 - 4x) + C.

9. Answer: (b) log_e4

Explanation: \(as\,\frac d{dx}log|4^x+x^4|=\frac1{4^x+x^4}.(4^x.log_e4+4x^3)\) \(=\frac{4x^3+log_e\,4.4^x}{4^x+x^4}\)

\(\Rightarrow\lambda=log_e\,4\)

10. Answer: (c) 3/2

Explanation: \(as\int\frac1{\sqrt{4-9x^2}}dx=\frac13\int\frac1{\sqrt{(\frac23)^2-x^2}}dx\)

\(=\frac13sin^{-1}(\frac{3x}2)+C\)

\(\Rightarrow a=\frac32.\)

11. Answer: (d) log (10^x + x¹⁰) + c.

Explanation: Given:

\(\int\frac{(10x^9+10x\,In\,10)dx}{(x^{10}+10^x)}\)

To evaluate

take \(y=x^{10}+10^x\)

\(\frac{dy}{dx}=10x^9+10x\,In\,10\)

\(dy=(10x^9+10^x\,In\,10)\;dx\)

\(\therefore\) \(I=\int\frac{dy}y=In\,y\)

Hence, the correct answer is \(In|(x^{10}+10^x)|\)

12. Answer: (b) tan (xe^x) + c

Explanation: Let \(I=\int\frac{e^x(1+x)}{cos^2(xe^x)}dx\)

Put \(x.e^x\) = t

Diff.w.r.t..x

\(\therefore\) \(x.e^x+e^x.1=\frac{dt}{dx}\)

\(\therefore\) \(e^x(1+x)dx=dt\)

\(\therefore I=\int\frac{dt}{cos^2t}=\int sec^2tdt\)

\(\therefore\) tan t + c

\(=tan(x.e^x)+e\)

13. Answer: (a) 1/6

Explanation: Given :

\(\therefore\) \(\int\frac1{(x^2+4)(x^2+9)}dx\) \(=A\,tan^{-1}\frac x2+B\,tan^{-1}\frac x3+C\;\;...(i)\)

\(\frac1{AB}=\frac1{B-A}(\frac1{A}-\frac1{B})\)

\(\therefore\) \(\int\frac1{(x^2+4)(x^2+9)}dx\) \(=\int\frac1{5}(\frac1{x^2+4}-\frac1{x^2+9})dx\)

\(=\frac1{5}\left[\frac1{2}tan^{-1}\frac x{2}-\frac1{3}tan^{-1}\frac x{3}\right]+C\)

\(=\frac1{10}tan^{-1}\frac x{2}-\frac1{15}tan^{-1}\frac x{3}+C\)

Comparing above equation with (i) we get

A = \(\frac1{10}\) and B = \(-\frac1{15}\)

\(\therefore\) A - B \(=\frac{1}{10}+\frac{1}{15}=\frac{5}{30}=\frac{1}{6}\)

14. Answer: (a) e^xtan^-1x+C

Explanation: \(\int\frac{e^x(x^2tan^{-1}x+tan^{-1}x+1)}{x^2+1}dx\) \(=\int\frac{e^x[(x^2+1)\,tan^{-1}x)+1]}{x^2+1}dx\)

\(=\int e^x(tan^{-1}x+\frac{1}{1+x^2})dx\) \(=e^xtan^{-1}x+c\)

Note : \(\int\) \(e^x\)[f(x) + f'(x)] dx = \(e^x\) f(x) + c

Here f(x) = \(tan^{-1}x\)

15. Answer: (c) a = -1/10, b = 2/5 

Explanation: \(I=\int\frac{dx}{(x+2)(x^2+1)}\)

\(\frac{1}{(x+2)(x^2+1)}=\frac A{x+2}+\frac{Bx+C}{x^2+1}\)

\(\Rightarrow\) 1 = A(\(x^2\)+1) + (Bx+C) (x+2)

\(\Rightarrow\) 1 = (A+B)\(x^2\) + (2B+C)x + A + 2C

Comparing coefficients, we get

A+B = 0, A+2C = 1, 2B+C = 0

Solving we get A = \(\frac15\), B = \(-\frac15\) and C = \(\frac25\)

\(\therefore\) \(\int\frac{dx}{(x+2)(x^2+1)}\)

\(=\frac15\int\frac1{x+2}dx+\int\cfrac{-\frac{1}{5}x+\frac{2}{5}}{x^2+1}dx\)

\(=\frac15\int\frac1{x+2}dx-\frac1{10}\int\frac{2x}{1+x^2}dx+\frac15\int\frac{2}{1+x^2}dx\)

\(=\frac15log|x+2|-\frac1{10}log|1+x^2|\) \(+\frac25tan^{-1}x+C\)

\(=a\,log|1+x^2|+b\,tan^{-1}x\) \(+\frac15\,log|x+2|+C\) (given)

\(\therefore\) \(a=\frac{-1}{10},b=\frac25\)

16. Answer: (a) 1/3 e^{x^3}+C

Explanation: \(\int x^2e^xdx\)

Let \(x^3\) = t

\(\Rightarrow\) \(3x^2=\frac{dt}{dx}\)

\(\Rightarrow dx=dt/3x^2\)

\(\therefore\) \(\int x^2\times e^t\times\frac{dt}{3x^2}\)

\(=\frac13\int e^tdt\)

\(=\frac13e^t+c\)

\(=\frac13\times e^{x^3}+c\)

17. Answer: (b) e^x sec x + c

Explanation: Let \(I=\int e^x\) sec x (1+tan x)dx

\(=\int e^x\) (sec x + sec x tan x)dx

Also, let sec x = f(x)

\(\Rightarrow\) sec x tan x = f'(x)

We know that,

\(\int e^x\) {f(x) + f'(x)} = \(e^x\) f(x) + C

\(\therefore\) I = \(e^x\) sec x + C

18. Answer: (a) (x + 1)tan^-1 √x – √x + c

Explanation: \(\int tan^{-1}\;\sqrt xdx\)

Apply integration by parts

\(x\int tan^{-1}\;\sqrt x-\int\frac x{2(1+x)\sqrt x}dx\)

\(x\int tan^{-1}\;\sqrt x-\frac12\int\frac{\sqrt x}{1+x}dx\)

Let \(\sqrt x\) = t

\(\frac{dx}{2\sqrt x}=dt\)

dx = 2tdt

\(=x\,tan^{-1}\;\sqrt x-\int\frac{t^2}{1+t^2}dt\)

\(=x\,tan^{-1}\;\sqrt x-\int dt+\int\frac{1}{1+t^2}dt\)

\(x\,tan^{-1}\;\sqrt x-\sqrt x+tan^{-1}\sqrt x+c\)

\(=(x-1)tan^{-1}\sqrt x-\sqrt x+c\)

19. Answer: (a) e^x cosx + C

Explanation: since \(\int e^x\) (f(x) + f'(x))dx = \(e^x\) f(x) + C

Here f(x) = cos x, f'(x) = -sin x. 

So, \(\int e^x\) (cos x - sin x)dx = \(e^x\) cos x + C

20.  Answer: (d) -2 ≤ a ≤ 4

Explanation: as \(\int_0^axdx\) ≤ a + 4
⇒ a²/2 ≤ a + 4
⇒ a² – 2a — 8 ≤ 0
⇒ (a – 1)² ≤ (3)²
⇒ -3 ≤ a – 1 ≤ 3
⇒ -2 ≤ a ≤ 4

Click here to Practice MCQ Question for Integrals Class 12

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Click here to start practice: - Class 12 Integrals Mock Test

∫e^-3x(cosx – 3sinx) dx

∫ e^-3x (-3sinx + cosx) 

f(x) = sinx,f'(x) = cosx 

∫e^-3x(-3sinx + cosx) 

= e^-3xsinx + C

Answer is (B)

= ∫ sec²(xe^x). e^x(1 + x)dx

= tan (xe^x) + C

∫sec²(7 - 4x)dx

= \(\frac{tan(7 - 4x)}{-4}\)+ c

(A) is the correct answer 

since ∫e^x [ f ( x) + f '( x)] dx = e^x f ( x) + C . Here

f (x) = cosx, f′ (x) = – sin x.

∫sinx sin (cosx)dx

cosx = t ⇒ – sinx . dx = dt 

∫- sin t. dt = cos t + C 

= cos (cos x) + C

∫e^x (sin x + cosx) dx 

put f(x) = sin x 

f’(x) = cos x 

= ∫e^x (f(x) + f’(x)) dx 

= e^xf'(x) + C 

=∫e^x (sinx + cosx) dx 

= e^x sin x + C

∫(2sec²x – 3secxtanx)dx

= 2 tan x – 3 sec x + C

∫(2x – 3cosx + e^x)dx

= \(\frac{2x^2}{2} \) + 3sinx + e^x

= x² - 3sinx + e^x + c

∫sec x (secx + tanx)dx 

= ∫(sec²x+sec x tanx)dx 

= tanx + secx + C

∫x²(1 - \(\frac{1}{x^2}\))dx

= ∫(x² - 1)dx

= \(\frac{x^3}{3}\)- x + C

∫(ax² + bx + c)dx

= a\(\frac{x^3}{3}\)+ b\(\frac{x^2}{2} + cx\) + d

Answer is f (x).

1. e^x + c

2. I = ∫e^x sinxdx = sinx.e^x – ∫cos x.e^xdx

= sin x.e^x – (cos x.e^x – ∫(- sin x).e^x dx)

= sinx.e^x – cosxe^x – ∫sinx.e^xdx

= sin x.e^x – cos xe^x – I

2I = sin x.e^x – cos xe^x

I = \(\frac{1}{2}\)e^x(sinx – cosx) + c.

∫(4e^3x + 1)dx = \(\frac{4e^{3x}}{3} + x\)

= \(\frac{4}{3}\)e^3x + x + C

∫(2x² + e^x)dx

= 2 x \(\frac{x^3}{3}\)+ e^x + C

\(\int\frac{x^2(x - 1) + 1(x - 1)}{(x - 1)}\)

= ∫(x² + 1)dx

= \(\frac{x^3}{3}\)+ x + c

∫tan⁴xdx

= ∫(sec²x - 1)tan²x.dx

= ∫sec²x.tan²x - ∫tan²x.dx

= ∫(tanx)²sec²x - ∫(sec²x - 1)dx

\(= \frac{tan^3x}{3} - tanx + x + c\)