Anand Classes provides a free downloadable PDF for NCERT Solutions – Class 12 Maths Chapter 7 (Integrals), Exercise 7.6 (Set-1), which covers the method of integration by parts. This study material aligns exactly with the latest CBSE/NCERT syllabus and is explained step-by-step by expert faculty to boost your preparation for board exams and competitive tests. Click the print button to download study material and notes.
NCERT Question 1: Evaluate the integral
$$\int x\sin x\; dx$$
Solution
Let
$$f(x)=\int x\sin x\; dx$$
Choose $x$ as the first function and $\sin x$ as the second function. Using integration by parts,
$$\int x\sin x\; dx = x\int \sin x\; dx – \int \left(\frac{d(x)}{dx}\right)\left(\int \sin x\; dx\right)\; dx$$
Compute the inner integral:
$$\int \sin x\; dx = -\cos x$$
Substituting,
$$f(x)= x(-\cos x) – \int 1(-\cos x)\; dx$$
This simplifies to
$$f(x)= -x\cos x + \int \cos x\; dx$$
Evaluate the remaining integral:
$$\int \cos x\; dx = \sin x$$
Hence,
$$f(x)= -x\cos x + \sin x + C$$
Final Answer
$$\boxed{-x\cos x + \sin x + C}$$
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NCERT Question 2: Evaluate the integral
$$\int x\sin 3x\; dx$$
Solution
Let
$$f(x)=\int x\sin 3x\; dx$$
Take $x$ as the first function and $\sin 3x$ as the second function. Using integration by parts,
$$\int x\sin 3x\; dx = x\int \sin 3x\; dx – \int \left(\frac{d(x)}{dx}\right)\left(\int \sin 3x\; dx\right)\; dx$$
Compute the inner integral:
$$\int \sin 3x\; dx = -\frac{\cos 3x}{3}$$
Substituting,
$$f(x)= x\left(-\frac{\cos 3x}{3}\right) – \int 1\left(-\frac{\cos 3x}{3}\right)\; dx$$
This becomes
$$f(x)= -\frac{x\cos 3x}{3} + \frac{1}{3}\int \cos 3x\; dx$$
Now evaluate the remaining integral:
$$\int \cos 3x\; dx = \frac{\sin 3x}{3}$$
Substitute back:
$$f(x)= -\frac{x\cos 3x}{3} + \frac{1}{9}\sin 3x + C$$
Final Answer
$$\boxed{-\frac{x\cos 3x}{3} + \frac{\sin 3x}{9} + C}$$
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NCERT Question 3: Evaluate the integral
$$\int x^{2} e^{x}\; dx$$
Solution
Let
$$f(x)=\int x^{2} e^{x}\; dx$$
Choose $x^{2}$ as the first function and $e^{x}$ as the second function. Using integration by parts,
$$\int x^{2} e^{x}\; dx = x^{2}\int e^{x}\; dx – \int \left(\frac{d(x^{2})}{dx}\right)\left(\int e^{x}\; dx\right)\; dx$$
Evaluate the inner integral:
$$\int e^{x}\; dx = e^{x}$$
Substitute:
$$f(x)= x^{2} e^{x} – \int 2x e^{x}\; dx$$
So,
$$f(x)= x^{2} e^{x} – 2\int x e^{x}\; dx$$
Now evaluate $\int x e^{x}\; dx$ by integration by parts again.
Let $u=x$ and $dv=e^{x}\; dx$, then
$$\int x e^{x}\; dx = x e^{x} – \int e^{x}\; dx$$
$$= x e^{x} – e^{x}$$
Substitute back:
$$f(x)= x^{2} e^{x} – 2\left(x e^{x} – e^{x}\right)$$
Simplify:
$$f(x)= x^{2} e^{x} – 2x e^{x} + 2 e^{x} + C$$
Factor out $e^{x}$:
$$f(x)= e^{x}(x^{2} – 2x + 2) + C$$
Final Answer
$$\boxed{e^{x}(x^{2}-2x+2) + C}$$
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NCERT Question 4: Evaluate the integral
$$\int x\log x\; dx$$
Solution
Let
$$f(x)=\int x\log x\; dx$$
Take $\log x$ as the first function and $x$ as the second function. Using integration by parts,
$$\int x\log x\; dx = \log x\int x\; dx – \int \left(\frac{d(\log x)}{dx}\right)\left(\int x\; dx\right)\; dx$$
Evaluate the inner integral:
$$\int x\; dx = \frac{x^{2}}{2}$$
Also,
$$\frac{d(\log x)}{dx} = \frac{1}{x}$$
Substitute these into the expression:
$$f(x)= \log x\left(\frac{x^{2}}{2}\right) – \int \frac{1}{x}\left(\frac{x^{2}}{2}\right)\; dx$$
Simplify inside the integral:
$$\frac{x^{2}}{2}\cdot\frac{1}{x}=\frac{x}{2}$$
So,
$$f(x)= \frac{x^{2}\log x}{2} – \int \frac{x}{2}\; dx$$
Integrate the remaining term:
$$\int \frac{x}{2}\; dx = \frac{x^{2}}{4}$$
Thus,
$$f(x)= \frac{x^{2}\log x}{2} – \frac{x^{2}}{4} + C$$
Final Answer
$$\boxed{\frac{x^{2}\log x}{2} – \frac{x^{2}}{4} + C}$$
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NCERT Question 5: Evaluate the integral
$$\int x\log 2x\; dx$$
Solution
Let
$$f(x)=\int x\log 2x\; dx$$
Take $\log 2x$ as the first function and $x$ as the second function. Using integration by parts,
$$\int x\log 2x\; dx = \log 2x\int x\; dx – \int \left(\frac{d(\log 2x)}{dx}\right)\left(\int x\; dx\right)\; dx$$
Evaluate the inner integral:
$$\int x\; dx = \frac{x^{2}}{2}$$
Differentiate the first function:
$$\frac{d(\log 2x)}{dx} = \frac{1}{2x}\times2=\frac{1}{x}$$
Substituting these results:
$$f(x)= \log 2x\left(\frac{x^{2}}{2}\right) – \int \frac{1}{x}\left(\frac{x^{2}}{2}\right)\; dx$$
Simplify inside the integral:
$$\frac{x^{2}}{2}\cdot\frac{1}{x}=\frac{x}{2}$$
Thus,
$$f(x)= \frac{x^{2}\log 2x}{2} – \int \frac{x}{2}\; dx$$
Integrate the remaining expression:
$$\int \frac{x}{2}\; dx = \frac{x^{2}}{4}$$
Therefore,
$$f(x)= \frac{x^{2}\log 2x}{2} \;-\; \frac{x^{2}}{4} + C$$
Final Answer
$$\boxed{\frac{x^{2}\log 2x}{2} \;-\; \frac{x^{2}}{4} + C}$$
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NCERT Question 6: Evaluate the integral
$$\int x^{2}\log x\; dx$$
Solution
Let
$$f(x)=\int x^{2}\log x\; dx$$
Take $\log x$ as the first function and $x^{2}$ as the second function. Using integration by parts,
$$\int x^{2}\log x\; dx = \log x\int x^{2}\; dx \;-\; \int \left(\frac{d(\log x)}{dx}\right)\left(\int x^{2}\; dx\right)\; dx$$
Compute the inner integral:
$$\int x^{2}\; dx = \frac{x^{3}}{3}$$
Differentiate $\log x$:
$$\frac{d(\log x)}{dx}=\frac{1}{x}$$
Substitute both results:
$$f(x)= \log x\left(\frac{x^{3}}{3}\right) – \int \frac{1}{x}\left(\frac{x^{3}}{3}\right)\; dx$$
Simplify the integrand:
$$\frac{x^{3}}{3}\cdot\frac{1}{x}=\frac{x^{2}}{3}$$
So,
$$f(x)= \frac{x^{3}\log x}{3} \;-\; \int \frac{x^{2}}{3}\; dx$$
Integrate the remaining term:
$$\int \frac{x^{2}}{3}\; dx = \frac{x^{3}}{9}$$
Thus,
$$f(x)= \frac{x^{3}\log x}{3} \;-\; \frac{x^{3}}{9} + C$$
Final Answer
$$\boxed{\frac{x^{3}\log x}{3} \;-\; \frac{x^{3}}{9} + C}$$
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NCERT Question 7: Evaluate the integral
$$\int x\sin^{-1}x\;dx$$
Solution
Let
$$f(x)=\int x\sin^{-1}x\;dx$$
Choose
$$u=\sin^{-1}x$$
$$dv=x\;dx$$
Then
$$du=\frac{1}{\sqrt{1-x^{2}}}\;dx$$
$$v=\frac{x^{2}}{2}$$
Using integration by parts :
$$f(x)=uv-\int v\;du$$
So
$$f(x)=\frac{x^{2}}{2}\sin^{-1}x-\int \frac{x^{2}}{2}\cdot\frac{1}{\sqrt{1-x^{2}}}\;dx$$
Rewrite the integrand:
$$x^{2}=1-(1-x^{2})$$
Thus
$$\frac{x^{2}}{\sqrt{1-x^{2}}}=\frac{1-(1-x^{2})}{\sqrt{1-x^{2}}}
=\frac{1}{\sqrt{1-x^{2}}}-\sqrt{1-x^{2}}$$
So the integral becomes
$$\int \frac{x^{2}}{\sqrt{1-x^{2}}}\;dx
=\int \frac{1}{\sqrt{1-x^{2}}}\;dx-\int \sqrt{1-x^{2}}\;dx$$
Now use the standard result
$$\int \sqrt{1-x^{2}}\;dx=\frac{x}{2}\sqrt{1-x^{2}}+\frac{1}{2}\sin^{-1}x$$
Hence
$$\int \frac{x^{2}}{\sqrt{1-x^{2}}}\;dx
=\sin^{-1}x-\left(\frac{x}{2}\sqrt{1-x^{2}}+\frac{1}{2}\sin^{-1}x\right)$$
This simplifies to
$$\frac{1}{2}\sin^{-1}x-\frac{x}{2}\sqrt{1-x^{2}}$$
Substitute back into $f(x)$:
$$f(x)=\frac{x^{2}}{2}\sin^{-1}x-\frac{1}{2}\left(\frac{1}{2}\sin^{-1}x-\frac{x}{2}\sqrt{1-x^{2}}\right)+C$$
Break into smaller parts:
$$f(x)=\frac{x^{2}}{2}\sin^{-1}x-\frac{1}{4}\sin^{-1}x+\frac{x}{4}\sqrt{1-x^{2}}+C$$
Combine the first two terms:
$$f(x)=\left(\frac{x^{2}}{2}-\frac{1}{4}\right)\sin^{-1}x+\frac{x}{4}\sqrt{1-x^{2}}+C$$
Final Answer
$$\boxed{\frac{1}{4}(2x^{2}-1)\sin^{-1}x+\frac{x}{4}\sqrt{1-x^{2}}+C}$$
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NCERT Question 8: Evaluate the integral
$$\int x\tan^{-1}x\;dx$$
Solution
Let
$$f(x)=\int x\tan^{-1}x\;dx$$
Choose
$$u=\tan^{-1}x$$
$$dv=x\;dx$$
Then
$$du=\frac{1}{1+x^{2}}\;dx$$
$$v=\frac{x^{2}}{2}$$
By integration by parts\;
$$f(x)=uv-\int v\;du$$
So
$$f(x)=\frac{x^{2}}{2}\tan^{-1}x-\int \frac{x^{2}}{2}\cdot\frac{1}{1+x^{2}}\;dx$$
Simplify the integrand:
$$\frac{x^{2}}{1+x^{2}}=1-\frac{1}{1+x^{2}}$$
Thus the integral becomes
$$\int \frac{x^{2}}{1+x^{2}}\;dx=\int\left(1-\frac{1}{1+x^{2}}\right)\;dx$$
Evaluate the simpler integrals:
$$\int 1\;dx=x$$
$$\int \frac{1}{1+x^{2}}\;dx=\tan^{-1}x$$
Therefore
$$\int \frac{x^{2}}{1+x^{2}}\;dx=x-\tan^{-1}x$$
Substitute back into $f(x)$:
$$f(x)=\frac{x^{2}}{2}\tan^{-1}x-\frac{1}{2}\bigl(x-\tan^{-1}x\bigr)+C$$
Break into smaller parts:
$$f(x)=\frac{x^{2}}{2}\tan^{-1}x-\frac{x}{2}+\frac{1}{2}\tan^{-1}x+C$$
Combine the $\tan^{-1}x$ terms if desired:
$$f(x)=\frac{1}{2}\bigl(x^{2}+1\bigr)\tan^{-1}x-\frac{x}{2}+C$$
Final Answer
$$\boxed{\displaystyle \int x\tan^{-1}x\;dx=\frac{1}{2}\bigl(x^{2}+1\bigr)\tan^{-1}x-\frac{x}{2}+C}$$
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NCERT Question 9: Evaluate the integral
$$\int x\cos^{-1}x\;dx$$
Solution
Let
$$f(x)=\int x\cos^{-1}x\;dx$$
Choose
$$u=\cos^{-1}x$$
$$dv=x\;dx$$
Then
$$du=-\frac{1}{\sqrt{1-x^{2}}}\;dx$$
$$v=\frac{x^{2}}{2}$$
By integration by parts\;
$$f(x)=uv-\int v\;du$$
So
$$f(x)=\frac{x^{2}}{2}\cos^{-1}x+\frac{1}{2}\int\frac{x^{2}}{\sqrt{1-x^{2}}}\;dx$$
Write the integrand in simpler pieces:
$$x^{2}=1-(1-x^{2})$$
Hence
$$\frac{x^{2}}{\sqrt{1-x^{2}}}=\frac{1-(1-x^{2})}{\sqrt{1-x^{2}}}
=\frac{1}{\sqrt{1-x^{2}}}-\sqrt{1-x^{2}}$$
So the integral becomes
$$\int\frac{x^{2}}{\sqrt{1-x^{2}}}\;dx
=\int\frac{1}{\sqrt{1-x^{2}}}\;dx-\int\sqrt{1-x^{2}}\;dx$$
Use the standard antiderivative
$$\int\sqrt{1-x^{2}}\;dx=\frac{x}{2}\sqrt{1-x^{2}}+\frac{1}{2}\sin^{-1}x$$
Therefore
$$\int\frac{x^{2}}{\sqrt{1-x^{2}}}\;dx
=\sin^{-1}x-\left(\frac{x}{2}\sqrt{1-x^{2}}+\frac{1}{2}\sin^{-1}x\right)$$
This simplifies to
$$\int\frac{x^{2}}{\sqrt{1-x^{2}}}\;dx
=\frac{1}{2}\sin^{-1}x-\frac{x}{2}\sqrt{1-x^{2}}$$
Substitute back into $f(x)$:
$$f(x)=\frac{x^{2}}{2}\cos^{-1}x+\frac{1}{2}\left(\frac{1}{2}\sin^{-1}x-\frac{x}{2}\sqrt{1-x^{2}}\right)+C$$
Split the terms and simplify:
$$f(x)=\frac{x^{2}}{2}\cos^{-1}x+\frac{1}{4}\sin^{-1}x-\frac{x}{4}\sqrt{1-x^{2}}+C$$
Convert $\sin^{-1}x$ to $\cos^{-1}x$ where useful using $\sin^{-1}x=\tfrac{\pi}{2}-\cos^{-1}x$ and absorb constants into $C$ to obtain the clean final form:
$$\boxed{\displaystyle f(x)=\frac{1}{4}\bigl(2x^{2}-1\bigr)\cos^{-1}x-\frac{x}{4}\sqrt{1-x^{2}}+C}$$
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NCERT Question 10: Evaluate the integral
$$\int (\sin^{-1}x)^{2}\;dx$$
Solution
Let
$$f(x)=\int (\sin^{-1}x)^{2}\;dx$$
Use integration by parts with
$$u=(\sin^{-1}x)^{2}$$
$$dv=dx$$
Then
$$du=2\sin^{-1}x\cdot\frac{1}{\sqrt{1-x^{2}}}\;dx$$
$$v=x$$
By parts :
$$f(x)=uv-\int v\;du$$
So
$$f(x)=x(\sin^{-1}x)^{2}-\int x\cdot 2\sin^{-1}x\cdot\frac{1}{\sqrt{1-x^{2}}}\;dx$$
Write the remaining integral as $2I$ where
$$I=\int \frac{x\sin^{-1}x}{\sqrt{1-x^{2}}}\;dx$$
Use the substitution $x=\sin t$ so that $\sin^{-1}x=t$ and $dx=\cos t\;dt$ and $\sqrt{1-x^{2}}=\cos t$. Then the integrand becomes $t\sin t$ and
$$I=\int t\sin t\;dt$$
Integrate $I$ by parts with $u=t$, $dv=\sin t \;dt$. Then $du=dt$, $v=-\cos t$ and
$$I=-t\cos t+\int \cos t\;dt=-t\cos t+\sin t + C$$
Return to $x$ using $t=\sin^{-1}x$, $\cos t=\sqrt{1-x^{2}}$, $\sin t=x$:
$$I=-\sin^{-1}x\sqrt{1-x^{2}}+x + C$$
Therefore the original integral is
$$ f(x) =x(\sin^{-1}x)^{2}-2I + C$$
$$f(x) =x(\sin^{-1}x)^{2}-2\bigl(-\sin^{-1}x\sqrt{1-x^{2}}+x\bigr)+C$$
Split into smaller pieces and simplify:
$$f(x)=x(\sin^{-1}x)^{2}+2\sin^{-1}x\sqrt{1-x^{2}}-2x + C$$
Final Answer
$$\boxed{\;\int (\sin^{-1}x)^{2}\;dx = x(\sin^{-1}x)^{2}+2\sin^{-1}x\sqrt{1-x^{2}}-2x + C\;}$$
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