Linear First Order Differential Equations - Mathematics Multiple Choice Questions 2

This set of Mathematics MCQs for Engineering Entrance Exams 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) \(\frac{3x}{4}-\frac{2sin⁡2x}{2}+C\)
View Answer

Answer: b
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 tan^3⁡x sec^2⁡x.
a) 2 tan⁴⁡x+C
b) 4 cot⁴⁡x+C
c) 2 tan^3⁡x+C
d) tan^4⁡x+C
View Answer

Answer: a
Explanation: To find: \(\int 8 \,tan^3⁡x \,sec^2⁡x \,dx\)
Let tan⁡x=t
sec^2⁡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
View Answer

Answer: 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.

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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) \(\frac{x}{2}+\frac{cos⁡(16x+5)}{32}+C\)
View Answer

Answer: 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
View Answer

Answer: b
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

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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⁡xe^x+C
d) -cos^2⁡xe^-x+C
View Answer

Answer: b
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
View Answer

Answer: d
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) –\(\frac{3}{4} \,(cos⁡(x+2))+\frac{1}{12} \,sin⁡(x+2)+C\)
View Answer

Answer: 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
View Answer

Answer: a
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) –\(\frac{3}{2} cos⁡x+\frac{cos⁡3x}{6}+C\)
View Answer

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

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