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Class 12 Maths MCQ Questions of Integrals with Answers?

Answer 1

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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 2

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

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

\(\Rightarrow a=\frac32.\)

Answer 3

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

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