Class 12 Maths MCQ – Methods of Integration-2

This set of Class 12 Maths Chapter 7 Multiple Choice Questions & Answers (MCQs) focuses on “Methods of Integration-2”.

1. Integrate 2 sin2⁡x+cos2x.
a) $$\frac{3x}{2}+\frac{sin⁡2x}{4}+C$$
b) $$\frac{3x}{2}-\frac{sin⁡2x}{4}+C$$
c) $$\frac{x}{2}+\frac{sin⁡2x}{4}+C$$
d) $$\frac{3x}{4}-\frac{2sin⁡2x}{2}+C$$

Explanation: $$\int \,2 \,sin^2⁡x +cos^2⁡x=\int sin^2⁡x+sin^2⁡x+cos^2⁡x dx$$
=$$\int sin^2⁡x+1 dx$$
=$$\int sin^2⁡x dx+\int 1 dx$$
=$$\int \frac{1-cos⁡2x}{2} dx+\int 1 dx$$
=$$\frac{x}{2}-\frac{sin⁡2x}{4}+x$$
=$$\frac{3x}{2}-\frac{sin⁡2x}{4}+C$$

2. Integrate 8 tan3⁡x sec2⁡x.
a) 2 tan4⁡x+C
b) 4 cot4⁡x+C
c) 2 tan3⁡x+C
d) tan4⁡x+C

Explanation: To find: $$\int 8 \,tan^3⁡x \,sec^2⁡x \,dx$$
Let tan⁡x=t
sec2⁡x dx=dt
∴$$\int 8 \,tan^3⁡x \,sec^2⁡x \,dx=\int 8 \,t^3 \,dt=\frac{8t^4}{4}=2t^4$$
Replacing t with tan⁡x, we get
$$\int 8 tan^3⁡x sec^2⁡x dx=2 tan^4⁡x+C$$

3. Find the integral of $$\frac{cos^2⁡x-sin^2⁡x}{7 cos^2⁡x sin^2⁡x}$$.
a) –$$\frac{1}{7}$$ (cot⁡x-tan⁡x)+C
b) –$$\frac{1}{7}$$ (cot⁡x-2 tan⁡x)+C
c) –$$\frac{1}{7}$$ (cot⁡x+tan⁡x)+C
d) –$$\frac{1}{7}$$ (2 cot⁡x+3 tan⁡x)+C

Explanation: To find: $$\int \frac{cos^2⁡x-sin^2⁡x}{7 \,cos^2⁡x \,sin^2⁡x} dx$$
$$\int \frac{cos^2⁡x-sin^2⁡x}{7 \,cos^2⁡x \,sin^2⁡x} dx=\frac{1}{7} \int \frac{1}{sin^2⁡x}-\frac{1}{cos^2⁡x} dx$$
=$$\frac{1}{7} \int cosec^2 x-sec^2⁡x dx$$
=$$\frac{1}{7}$$ (-cot⁡x-tan⁡x)+C
=-$$\frac{1}{7}$$ (cot⁡x+tan⁡x)+C.

4. Find $$\int sin^2⁡(8x+5) dx$$
a) $$\frac{x}{4}+\frac{sin⁡(16x+10)}{32}+C$$
b) $$\frac{x}{2}-\frac{cos⁡(16x+10)}{32}+C$$
c) $$\frac{x}{2}-\frac{sin⁡(16x+10)}{32}+C$$
d) $$\frac{x}{2}+\frac{cos⁡(16x+5)}{32}+C$$

Explanation: $$\int sin^2⁡(8x+5) dx=\int \frac{1-cos⁡2(8x+5)}{2} dx=\int \frac{1}{2} dx-\frac{1}{2} \int cos(16x+10)dx$$
=$$\frac{x}{2}-\frac{1}{2} (\frac{sin⁡(16x+10)}{16})=\frac{x}{2}-\frac{sin⁡(16x+10)}{32}+C$$

5. Find $$\int \frac{5 cos^2⁡x}{1+sin⁡x} dx$$.
a) -3(x+cos⁡x)+C
b) 5(x+cos⁡x)+C
c) 5(-x+sin⁡x)+C
d) 5(x-cos⁡x)+C

Explanation: $$\int \frac{5 cos^2⁡x}{1+sin⁡x} dx=\int \frac{5(1-sin^2⁡x)}{1+sin⁡x}=5\int \frac{(1+sin⁡x)(1-sin⁡x)}{(1+sin⁡x)} dx$$
=5∫ (1-sin⁡x)dx
=5(x-(-cos⁡x))=5(x+cos⁡x)+C

6. Find the integral of $$\frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})}$$.
a) cot⁡xe-x+C
b) -cot⁡xe-x+C
c) -cot⁡xex+C
d) -cos2⁡xe-x+C

Explanation: $$\int \frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})} dx$$
Let xe-x=t
Differentiating w.r.t x, we get
$$-xe^{-x}+e^{-x} dx=dt$$
e-x (1-x)dx=dt
$$\int \frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})} dx=\int \frac{dt}{sin^2⁡t}$$
=$$\int cosec^2 \,t \,dt$$
=-cot⁡t+C
Replacing t with xe-x, we get
$$\int \frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})} dx=-cot⁡xe^{-x}+C$$.

7. Integrate $$\frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2}$$.
a) -log⁡(1+2sin⁡2x)+C
b) $$\frac{1}{4}$$ log⁡(1-sin⁡2x)+C
c) –$$\frac{1}{4}$$ log⁡(1+cos⁡2x)+C
d) -log⁡(1-sin⁡2x)+C

Explanation: $$\int \frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2} dx=\int \frac{2 cos⁡2x}{cos^2⁡x+sin^2⁡x-sin⁡2x} \,dx \,(∵2 cos⁡x sin⁡x=sin⁡2x)$$
=$$\int \frac{2 cos⁡2x}{1-sin⁡2x} dx$$
Let 1-sin⁡2x=t
Differentiating w.r.t x, we get
-2 cos⁡2x dx=dt
2 cos⁡2x dx=-dt
$$\int \frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2} dx=-\int \frac{dt}{t}$$
=-log⁡t
Replacing t with 1-sin⁡2x, we get
∴$$\int \frac{2 cos⁡2x}{(cos⁡x-sin⁡x)^2}$$ dx=-log⁡(1-sin⁡2x)+C

8. Integrate sin3⁡(x+2).
a) $$\frac{3}{4} \,(sin⁡(x+2))+\frac{1}{12} \,cos⁡(3x+6)+C$$
b) –$$\frac{3}{4} \,(cos⁡(x+2))-\frac{1}{5} \,cos⁡(3x+6)+C$$
c) –$$\frac{3}{4} \,(cos⁡(x+2))+\frac{1}{12} \,cos⁡(3x+6)+C$$
d) –$$\frac{3}{4} \,(cos⁡(x+2))+\frac{1}{12} \,sin⁡(x+2)+C$$

Explanation: To find: ∫ 3 sin3⁡(x+2) dx
We know that, sin⁡3x=3 sin⁡x-4 sin3⁡x
∴sin3⁡⁡x=$$\frac{3 sin⁡x-sin⁡3x}{4}$$
sin3⁡(x+2)=$$\frac{(3 sin⁡(x+2)-sin⁡(3x+6))}{4}$$
$$\int sin^3⁡(x+2) \,dx=\frac{3}{4} \int sin⁡(x+2) \,dx-\frac{1}{4} \int \,sin⁡(3x+6) \,dx$$
=-$$\frac{3}{4} \,(cos⁡(x+2))+\frac{1}{12} \,cos⁡(3x+6)+C$$

9. Integrate 2x cos⁡(x2+3).
a) sin⁡(x2+3)+C
b) sin2⁡(x2+3)+C
c) cot⁡(x2+3)+C
d) -sin⁡(x2+3)+C

Explanation: ∫ 2x cos⁡(x2+3) dx
Let x2+3=t
Differentiating w.r.t x, we get
2x dx=dt
∫ 2x cos⁡(x2+3) dx=∫ cos⁡t dt
=sin⁡t+C
Replacing w.r.t x, we get
∴∫ 2x cos⁡(x2+3) dx=sin⁡(x2+3)+C

10. Find ∫ 2 sin3⁡x+1 dx
a) $$\frac{3}{2}-\frac{cos⁡3x}{6}+x+C$$
b) –$$\frac{3}{2} cos⁡x+\frac{cos⁡3x}{6}+x+C$$
c) –$$\frac{3}{2} cos⁡x-\frac{cos⁡3x}{6}-x+C$$
d) –$$\frac{3}{2} cos⁡x+\frac{cos⁡3x}{6}+C$$

Explanation: We know that, sin⁡3x=3 sin⁡x-4 sin3⁡x
∴sin3x=$$\frac{3 sin⁡x-sin⁡3x}{4}$$
$$\int 2 \,sin^3⁡x+1 \,dx=\int \frac{(3 sin⁡x-sin⁡3x)}{2} dx+\int dx$$
=$$\frac{3}{2} \int sin⁡x dx-\frac{1}{2} \int sin⁡3x dx+\int dx$$
=-$$\frac{3}{2} cos⁡x+\frac{cos⁡3x}{6}+x+C$$

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

To practice all chapters and topics of class 12 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.