This set of Class 12 Maths Chapter 7 Multiple Choice Questions & Answers (MCQs) 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
Explanation: By using the formula \(\int u.v \,dx=u\int \,v \,dx-\int u'(\int \,v \,dx)\) we get
\(\int xe^{2x} dx=x\int e^{2x} dx-\int (x)’\int e^{2x} \,dx\)
=\(\frac{xe^{2x}}{2}-\int 1.\frac{e^{2x}}{2} \,dx\)
=\(\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\)
=\(\frac{e^{2x}}{4} (2x-1)+C\)
2. Find ∫ 7 logx.x dx
a) \(\frac{7}{2} (logx-x)+C\)
b) –\(\frac{7}{2} (x^2 logx-x^3)+C\)
c) \(\frac{7}{2} (x^2 logx-x)+C\)
d) (x2 logx+x)+C
View Answer
Explanation: ∫ 7 logx.x dx=7∫ logx.x dx
Using ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx) , we get
7∫ logx.x dx=7(logx ∫ x dx-(logx)’∫ x dx)
=\(7\left (\frac{x^2 logx}{2}-\frac{1}{x}.\frac{x^2}{2}\right)\)
=\(\frac{7}{2} (x^2 logx-x)+C\)
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
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 \(\int \,x \,e^{2x} \,dx\)
\(\int \,x \,e^{2x} \,dx=x\int e^{2x} \,dx-\int (x)’ \int \,e^{2x} \,dx\)
=\(\frac{xe^{2x}}{2}-\int \frac{e^{2x}}{2} \,dx\)
=\(\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\).
∴\(\int \,(x^2+9) \,e^x \,dx=\frac{(x^2+9) e^{2x}}{2}+\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\)
=\(\frac{e^{2x}}{2}(x^2+9+x-\frac{1}{4})\)
=\(\frac{e^{2x}}{2}(x^2+x-\frac{35}{4})+C\).
4. Integrate ∫ logx2 dx
a) logx2 + x+C
b) x logx2 – 2x+C
c) x logx2 – 1+C
d) x logx2 + x+C
View Answer
Explanation: By using ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx)
∫ logx2.1 dx=logx2 ∫ dx-\(\int \frac{1}{x^2}.2x \int dx\)
=x logx2 – 2∫ 1/x.x dx
=x logx2 – 2∫ dx
=x logx2 – 2x+C
5. Integrate 2x sin2x.
a) \(\frac{sin2x}{2}\)+x cos2x+C
b) \(\frac{sin2x}{2}\)-cos2x+C
c) \(\frac{cos2x}{2}\)-x cos2x+C
d) \(\frac{sin2x}{2}\)-x cos2x+C
View Answer
Explanation: By using ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx)
∫ x tanx dx=2x ∫ sin2x dx-∫ (2x)’ ∫ sin2x dx
=2x (-\(\frac{cos2x}{2}\))+2\(\int \frac{cos2x}{2}\) dx
=-xcos 2x+∫ cos2x dx
=\(\frac{sin2x}{2}\)-x cos2x+C
6. Integrate 3 sec2x log(tanx) dx.
a) -log(tanx) (tanx-1)+C
b) log(tanx) (secx+1)+C
c) tanx (log(tanx)-1)+C
d) tanx (logsecx +1)+C
View Answer
Explanation: By using∫ u.v dx=u∫ v dx-∫ u'(∫ v dx), we get
∫ log(tanx) sec2x dx=log(tanx) ∫ sec2 xdx -∫ (logtanx)’∫ sec2x dx
=tanx log(tanx)-\(\int \frac{1}{tanx} sec^2x.tanx \,dx\)
=tan xlog(tanx)-∫ sec2x dx
=tan xlog(tanx)-tanx+C
=tanx (log(tanx)-1)+C
7. Find ∫ 10 logx.x2 dx
a) \(\frac{10x^3}{3} \left(x^3 logx-\frac{x^3}{3}\right)+C\)
b) \(\frac{10x^3}{3} \left(logx-\frac{x^3}{3}\right)+C\)
c) \(-\frac{10x^3}{3} \left(x^3 logx-\frac{x^3}{3}\right)+C\)
d) \(\left(x^3 logx-\frac{x^3}{3}\right)+C\)
View Answer
Explanation: ∫ 10 logx.x2 dx=10∫ logx.x2 dx
By using the formula, ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx), we get
\(10\int logx.x^2 \,dx=10(logx \int x^2 \,dx-\int (logx)’\int \,x^2 \,dx)\)
=\(10 \left(\frac{x^3 logx}{3}-(\int \frac{1}{x}.\frac{x^3}{3} \,dx)\right)+C\)
=\(10 \left(\frac{x^3 logx}{3}-\frac{1}{3} \frac{x^3}{3}\right)+C\)
=\(\frac{10x^3}{3} \left(x^3 logx-\frac{x^3}{3}\right)+C\).
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
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 sin3x.
a) –\(\frac{5}{3} \,x \,cos3x+\frac{5}{9} tan3x+C\)
b) \(\frac{5}{3} \,cos3x-\frac{5}{9} \,sin3x+C\)
c) \(x cos3x+\frac{5}{9} \,sin3x+C\)
d) –\(\frac{5}{3} \,x \,cos3x+\frac{5}{9} \,sin3x+C\)
View Answer
Explanation: By using the formula, ∫ u.v dx=u∫ v dx-∫ u’ (∫ v dx) ,we get
∫ 5x sin3x dx=5x∫ sin3x dx -∫ (5x)’∫ sin3x dx
=\(5x(-\frac{cos3x}{3})+\int \frac{5 cos3x}{3} dx\)
=-\(\frac{5}{3} x cos3x+\frac{5}{9} sin3x+C\)
10. Find ∫ sinx log(cosx) dx.
a) cosx (log(sinx)-1)+C
b) sinx (log(cosx)+1)+C
c) cosx (log(cosx)-1)+C
d) cosx (log(cosx)-1)+C
View Answer
Explanation: Let cosx=t
Differentiating w.r.t x, we get
-sinx dx=dt
sinx dx=-dt
∴∫ sinx log(cosx) dx=∫ -logt dt
Using ∫ u.v dx=u∫ v dx-∫ u’ (∫ v dx) , we get
∫ -logt dt=-logt ∫ 1 dt-∫ (-logt)’ ∫ 1 dt
=-t logt+∫ dt
=-t logt+t
=t(logt-1)
Replacing t with cosx, we get
∫ sinx log(cosx) dx=cosx (log(cosx)-1)+C
Sanfoundry Global Education & Learning Series – Mathematics – Class 12.
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