Mathematics Questions and Answers – Methods of Integration-2

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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: \(\int \,2 \,sin^2⁡x +cos^2⁡x=\int sin^2⁡x+sin^2⁡x+cos^2⁡x dx\)
=\(\int sin^2⁡x+1 dx\)
=\(\int sin^2⁡x dx+\int 1 dx\)
=\(\int \frac{1-cos⁡2x}{2} dx+\int 1 dx\)
=\(\frac{x}{2}-\frac{sin⁡2x}{4}+x\)
=\(\frac{3x}{2}-\frac{sin⁡2x}{4}+C\)
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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
View Answer

Answer: a
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
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⁡xex+C
d) -cos2⁡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
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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
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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=\(\frac{3 sin⁡x-sin⁡3x}{4}\)
\(\int 2 \,sin^3⁡x+1 \,dx=\int \frac{(3 sin⁡x-sin⁡3x)}{2} dx+\int dx\)
=\(\frac{3}{2} \int sin⁡x dx-\frac{1}{2} \int sin⁡3x dx+\int dx\)
=-\(\frac{3}{2} cos⁡x+\frac{cos⁡3x}{6}+x+C\)

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

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter