Mathematics Questions and Answers – Integration by Parts

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This set of Mathematics Quiz for IIT JEE Exam focuses on “Integration by Parts”.

1. Integrate xe2x.
a) \(\frac{e^{2x}}{4} (x-\frac{1}{4})+C\)
b) \(\frac{e^{2x}}{4} (2x-1)+C\)
c) \(\frac{e^{2x}}{2} (2x-1)+C\)
d) \(\frac{e^{2x}}{4} (x+1)+C\)
View Answer

Answer: b
Explanation: By using the formula \(\int u.v \,dx=u\int \,v \,dx-\int u'(\int \,v \,dx)\) we get
\(\int xe^{2x} dx=x\int e^{2x} dx-\int (x)’\int e^{2x} \,dx\)
=\(\frac{xe^{2x}}{2}-\int 1.\frac{e^{2x}}{2} \,dx\)
=\(\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\)
=\(\frac{e^{2x}}{4} (2x-1)+C\)
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2. Find ∫ 7 log⁡x.x dx
a) \(\frac{7}{2} (log⁡x-x)+C\)
b) –\(\frac{7}{2} (x^2 log⁡x-x^3)+C\)
c) \(\frac{7}{2} (x^2 log⁡x-x)+C\)
d) (x2 log⁡x+x)+C
View Answer

Answer: c
Explanation: ∫ 7 log⁡x.x dx=7∫ log⁡x.x dx
Using ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx) , we get
7∫ log⁡x.x dx=7(log⁡x ∫ x dx-(log⁡x)’∫ x dx)
=\(7\left (\frac{x^2 log⁡x}{2}-\frac{1}{x}.\frac{x^2}{2}\right)\)
=\(\frac{7}{2} (x^2 log⁡x-x)+C\)

3. Integrate \((x^2+9) e^{2x} dx\)
a) \(\frac{e^{x}}{2} (x^2+x-\frac{39}{4})+C\)
b) \(\frac{e^{2x}}{2} (x^2+x-\frac{35}{4})+C\)
c) \(\frac{e^{2x}}{2} (x^2+x-\frac{48}{4})+C\)
d) \(\frac{e^{x}}{2} (x^2+x-\frac{25}{4})+C\)
View Answer

Answer: b
Explanation: By using ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx), we get
\(\int (x^2+9) \,e^x \,dx=(x^2+9) \int e^{2x} \,dx-\int (x^2+9)’\int e^{2x} \,dx\)
=\(\frac{(x^2+9) \,e^{2x}}{2}-\int 2x \frac{e^{2x}}{2} \,dx\)
Again, applying integration by parts for integrating \(\int \,x \,e^{2x} \,dx\)
\(\int \,x \,e^{2x} \,dx=x\int e^{2x} \,dx-\int (x)’ \int \,e^{2x} \,dx\)
=\(\frac{xe^{2x}}{2}-\int \frac{e^{2x}}{2} \,dx\)
=\(\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\).
∴\(\int \,(x^2+9) \,e^x \,dx=\frac{(x^2+9) e^{2x}}{2}+\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\)
=\(\frac{e^{2x}}{2}(x^2+9+x-\frac{1}{4})\)
=\(\frac{e^{2x}}{2}(x^2+x-\frac{35}{4})+C\).
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4. Integrate ∫ log⁡x2 dx
a) log⁡x2 + x+C
b) x log⁡x2 – 2x+C
c) x log⁡x2 – 1+C
d) x log⁡x2 + x+C
View Answer

Answer: b
Explanation: By using ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx)
∫ log⁡x2.1 dx=log⁡x2 ∫ dx-\(\int \frac{1}{x^2}.2x \int dx\)
=x log⁡x2 – 2∫ 1/x.x dx
=x log⁡x2 – 2∫ dx
=x log⁡x2 – 2x+C

5. Integrate 2x sin⁡2x.
a) \(\frac{sin⁡2x}{2}\)+x cos⁡2x+C
b) \(\frac{sin⁡2x}{2}\)-cos⁡2x+C
c) \(\frac{cos⁡2x}{2}\)-x cos⁡2x+C
d) \(\frac{sin⁡2x}{2}\)-x cos⁡2x+C
View Answer

Answer: d
Explanation: By using ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx)
∫ x tanx dx=2x ∫ sin⁡2x dx-∫ (2x)’ ∫ sin⁡2x dx
=2x (-\(\frac{cos⁡2x}{2}\))+2\(\int \frac{cos⁡2x}{2}\) dx
=-xcos 2x+∫ cos⁡2x dx
=\(\frac{sin⁡2x}{2}\)-x cos⁡2x+C
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6. Integrate 3 sec2⁡x log⁡(tan⁡x) dx.
a) -log⁡(tan⁡x) (tan⁡x-1)+C
b) log⁡(tan⁡x) (sec⁡x+1)+C
c) tan⁡x (log⁡(tan⁡x)-1)+C
d) tan⁡x (log⁡sec⁡x +1)+C
View Answer

Answer: c
Explanation: By using∫ u.v dx=u∫ v dx-∫ u'(∫ v dx), we get
∫ log⁡(tan⁡x) sec2x dx=log⁡(tan⁡x) ∫ sec2 x⁡dx -∫ (log⁡tan⁡x)’∫ sec2x dx
=tan⁡x log⁡(tan⁡x)-\(\int \frac{1}{tan⁡x} sec^2⁡x.tan⁡x \,dx\)
=tan x⁡log⁡(tan⁡x)-∫ sec2⁡x dx
=tan x⁡log⁡(tan⁡x)-tan⁡x+C
=tan⁡x (log⁡(tan⁡x)-1)+C

7. Find ∫ 10 log⁡x.x2 dx
a) \(\frac{10x^3}{3} \left(x^3 log⁡x-\frac{x^3}{3}\right)+C\)
b) \(\frac{10x^3}{3} \left(log⁡x-\frac{x^3}{3}\right)+C\)
c) \(-\frac{10x^3}{3} \left(x^3 log⁡x-\frac{x^3}{3}\right)+C\)
d) \(\left(x^3 log⁡x-\frac{x^3}{3}\right)+C\)
View Answer

Answer: a
Explanation: ∫ 10 log⁡x.x2 dx=10∫ log⁡x.x2 dx
By using the formula, ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx), we get
\(10\int log⁡x.x^2 \,dx=10(log⁡x \int x^2 \,dx-\int (log⁡x)’\int \,x^2 \,dx)\)
=\(10 \left(\frac{x^3 log⁡x}{3}-(\int \frac{1}{x}.\frac{x^3}{3} \,dx)\right)+C\)
=\(10 \left(\frac{x^3 log⁡x}{3}-\frac{1}{3} \frac{x^3}{3}\right)+C\)
=\(\frac{10x^3}{3} \left(x^3 log⁡x-\frac{x^3}{3}\right)+C\).
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8. Find ∫ 2x3 ex2 dx.
a) -ex2 (x2+2)+C
b) ex2 (x2-1)+C
c) 2ex2 (x2+1)+C
d) ex2 (x-1)+C
View Answer

Answer: b
Explanation: Let x2=t
Differentiating w.r.t x, we get
2x dx=dt
∴∫ 2x2 ex2 dx=∫ tet dt
By using the formula, ∫ u.v dx=u∫ v dx-∫ u’ (∫ v dx) ,we get
∫ t et dt=t∫ et dt-∫ (t)’∫ et dt
=tet-∫ et dt
=tet-et=et (t-1)
Replacing t with x2, we get
∫ 2x3 ex2 dx=ex2 (x2-1)+C

9. Integrate 5x sin⁡3x.
a) –\(\frac{5}{3} \,x \,cos⁡3x+\frac{5}{9} tan⁡3x+C\)
b) \(\frac{5}{3} \,cos⁡3x-\frac{5}{9} \,sin⁡3x+C\)
c) \(x cos⁡3x+\frac{5}{9} \,sin⁡3x+C\)
d) –\(\frac{5}{3} \,x \,cos⁡3x+\frac{5}{9} \,sin⁡3x+C\)
View Answer

Answer: d
Explanation: By using the formula, ∫ u.v dx=u∫ v dx-∫ u’ (∫ v dx) ,we get
∫ 5x sin⁡3x dx=5x∫ sin⁡3x dx -∫ (5x)’∫ sin⁡3x dx
=\(5x(-\frac{cos⁡3x}{3})+\int \frac{5 cos⁡3x}{3} dx\)
=-\(\frac{5}{3} x cos⁡3x+\frac{5}{9} sin⁡3x+C\)
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10. Find ∫ sin⁡x log⁡(cos⁡x) dx.
a) cos⁡x (log⁡(sin⁡x)-1)+C
b) sin⁡x (log⁡(cos⁡x)+1)+C
c) cos⁡x (log⁡(cos⁡x)-1)+C
d) cos⁡x (log⁡(cos⁡x)-1)+C
View Answer

Answer: c
Explanation: Let cos⁡x=t
Differentiating w.r.t x, we get
-sin⁡x dx=dt
sin⁡x dx=-dt
∴∫ sin⁡x log⁡(cos⁡x) dx=∫ -log⁡t dt
Using ∫ u.v dx=u∫ v dx-∫ u’ (∫ v dx) , we get
∫ -log⁡t dt=-log⁡t ∫ 1 dt-∫ (-log⁡t)’ ∫ 1 dt
=-t log⁡t+∫ dt
=-t log⁡t+t
=t(log⁡t-1)
Replacing t with cos⁡x, we get
∫ sin⁡x log⁡(cos⁡x) dx=cos⁡x (log⁡(cos⁡x)-1)+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