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 lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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