Anand Classes provides comprehensive NCERT Solutions for Integrals Exercise 7.6 of Chapter 7 for Class 12 Math, designed to help students strengthen their understanding of integration concepts with clear, step-by-step explanations. This Set-2 PDF offers accurate solutions aligned with the latest NCERT curriculum, making it a valuable resource for exam preparation and concept revision. Click the print button to download study material and notes.
NCERT Question 11: Evaluate the integral
$$\int \frac{x\cos^{-1}x}{\sqrt{1-x^{2}}}\;dx$$
Solution
Let
$$f(x)=\int \frac{x\cos^{-1}x}{\sqrt{1-x^{2}}}\;dx$$
Set
$$t=\cos^{-1}x$$
so that
$$x=\cos t$$
and
$$dt=-\frac{1}{\sqrt{1-x^{2}}}\;dx$$
Hence
$$dx=-\sqrt{1-x^{2}}\;dt$$
Substitute into the integral:
$$f(x)=\int \frac{x\cdot t}{\sqrt{1-x^{2}}}\;dx
= \int \frac{\cos t\cdot t}{\sin t}\cdot\bigl(-\sin t\bigr)\;dt$$
This simplifies to
$$f(x)=-\int t\cos t\;dt$$
Integrate by parts with $u=t$ and $dv=\cos t\;dt$. Then
$$du=dt\quad\text{and}\quad v=\sin t$$
So
$$-\int t\cos t\;dt = -\bigl(t\sin t-\int \sin t\;dt\bigr)$$
Evaluate the remaining integral:
$$\int \sin t\;dt = -\cos t$$
Thus
$$-\int t\cos t\;dt = -t\sin t -\cos t + C$$
Return to $x$ using $t=\cos^{-1}x$ , $\sin t=\sqrt{1-x^{2}}$ and $\cos t=x$:
$$f(x) = -\cos^{-1}x \;\sqrt{1-x^{2}} – x + C$$
Final Answer
$$\boxed{\;-\sqrt{1-x^{2}}\cos^{-1}x \;-\; x + C\;}$$
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NCERT Question 12: Evaluate the integral
$$\int x\sec^{2}x\;dx$$
Solution
Let
$$f(x)=\int x\sec^{2}x\;dx$$
Choose
$$u=x$$
$$dv=\sec^{2}x\;dx$$
Then
$$du=dx$$
$$v=\tan x$$
By integration by parts :
$$f(x)=uv-\int v\;du$$
So
$$f(x)=x\tan x-\int \tan x\;dx$$
Evaluate the remaining integral:
$$\int \tan x\;dx=-\ln\lvert\cos x\rvert$$
Substitute back:
$$f(x)=x\tan x+\ln\lvert\cos x\rvert + C$$
Final Answer
$$\boxed{\;x\tan x+\ln\lvert\cos x\rvert + C\;}$$
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NCERT Question 13: Evaluate the integral
$$\int \tan^{-1}x\;dx$$
Solution
Let
$$f(x)=\int \tan^{-1}x\;dx$$
Choose
$$u=\tan^{-1}x$$
$$dv=dx$$
Then
$$du=\frac{1}{1+x^{2}}\;dx$$
$$v=x$$
Apply integration by parts:
$$f(x)=uv-\int v\;du$$
So,
$$f(x)=x\tan^{-1}x-\int x\cdot\frac{1}{1+x^{2}}\;dx$$
Rewrite the remaining integral:
$$\int \frac{x}{1+x^{2}}\;dx=\frac{1}{2}\int \frac{2x}{1+x^{2}}\;dx$$
This integrates to
$$\int \frac{x}{1+x^{2}}\;dx=\frac{1}{2}\ln(1+x^{2})$$
Substitute back:
$$f(x)=x\tan^{-1}x-\frac{1}{2}\ln(1+x^{2})+C$$
Final Answer
$$\boxed{x\tan^{-1}x-\frac{1}{2}\ln(1+x^{2})+C}$$
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NCERT Question 14: Evaluate the integral
$$\int x(\log x)^{2}\;dx$$
Solution
Let
$$f(x)=\int x(\log x)^{2}\;dx$$
Choose
$$u=(\log x)^{2}$$
$$dv=x\;dx$$
Then
$$du=2\log x\cdot \frac{1}{x}\;dx$$
$$v=\frac{x^{2}}{2}$$
Using integration by parts:
$$f(x)=uv-\int v\;du$$
So,
$$f(x)=\frac{x^{2}}{2}(\log x)^{2}-\int \frac{x^{2}}{2}\cdot \frac{2\log x}{x}\;dx$$
Simplify the integrand:
$$\frac{x^{2}}{2}\cdot \frac{2\log x}{x}=x\log x$$
Thus,
$$f(x)=\frac{x^{2}}{2}(\log x)^{2}-\int x\log x\;dx$$
Now integrate
$$\int x\log x\;dx$$
again by parts.
Choose
$$u=\log x$$
$$dv=x\;dx$$
Then
$$du=\frac{1}{x}\;dx$$
$$v=\frac{x^{2}}{2}$$
Apply integration by parts again:
$$\int x\log x\;dx=\frac{x^{2}}{2}\log x-\int \frac{x^{2}}{2}\cdot\frac{1}{x}\;dx$$
Simplify:
$$\int \frac{x^{2}}{2}\cdot \frac{1}{x}\;dx=\frac{1}{2}\int x\;dx=\frac{x^{2}}{4}$$
Thus,
$$\int x\log x\;dx=\frac{x^{2}}{2}\log x-\frac{x^{2}}{4}$$
Substitute back into $f(x)$:
$$f(x)=\frac{x^{2}}{2}(\log x)^{2}-\left(\frac{x^{2}}{2}\log x-\frac{x^{2}}{4}\right)+C$$
Distribute the minus sign:
$$f(x)=\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2}\log x+\frac{x^{2}}{4}+C$$
Final Answer
$$\boxed{\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2}\log x+\frac{x^{2}}{4}+C}$$
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NCERT Question 15: Evaluate the integral
$$\int (x^{2}+1)\log x\; dx$$
Solution
Let
$$f(x)=\int (x^{2}+1)\log x\; dx$$
Split the integral:
$$f(x)=\int x^{2}\log x\; dx+\int \log x\; dx$$
Write
$$f(x)=I_{1}+I_{2}$$
Compute $I_{1}=\displaystyle\int x^{2}\log x\; dx$ by parts.
Choose $u=\log x$ and $dv=x^{2}\;dx$. Then
$$du=\frac{1}{x}\;dx\qquad v=\frac{x^{3}}{3}$$
So
$$I_{1}=\frac{x^{3}}{3}\log x-\int \frac{x^{3}}{3}\cdot\frac{1}{x}\;dx$$
Simplify the integrand and integrate:
$$I_{1}=\frac{x^{3}}{3}\log x-\frac{1}{3}\int x^{2}\;dx$$
$$I_{1}=\frac{x^{3}}{3}\log x-\frac{1}{3}\cdot\frac{x^{3}}{3}+C_{1}$$
$$I_{1}=\frac{x^{3}}{3}\log x-\frac{x^{3}}{9}+C_{1}$$
Now compute $I_{2}=\displaystyle\int \log x\; dx$ by parts.
Choose $u=\log x$ and $dv=dx$. Then
$$du=\frac{1}{x}\;dx\qquad v=x$$
So
$$I_{2}=x\log x-\int x\cdot\frac{1}{x}\;dx$$
$$I_{2}=x\log x-\int 1\;dx$$
$$I_{2}=x\log x-x+C_{2}$$
Combine $I_{1}$ and $I_{2}$:
$$
f(x)=\left(\frac{x^{3}}{3}\log x-\frac{x^{3}}{9}\right)
+\left(x\log x-x\right)+C $$
$$ f(x) =\left(\frac{x^{3}}{3}+x\right)\log x-\frac{x^{3}}{9}-x+C $$
Final Answer
$$\boxed{\displaystyle \int (x^{2}+1)\log x\; dx=\Bigl(\frac{x^{3}}{3}+x\Bigr)\log x-\frac{x^{3}}{9}-x+C}$$
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NCERT Question 16: Evaluate the integral
$$\int e^{x}(\sin x+\cos x)\;dx$$
Solution
$$f(x)=\int e^{x}(\sin x+\cos x)\;dx$$
$$g(x)=\sin x$$
$$g'(x)=\cos x$$
$$f(x)=\int e^{x}{\left[g(x)+g'(x)\right]}\;dx$$
Note : Important Property
$$\int e^{x}\;{[\;g(x)+g'(x)}]\;dx=e^{x}g(x)+C$$
$$f(x)=\int e^{x}(\sin x+\cos x)\;dx = e^{x}\sin x+C$$
Final Answer
$$\boxed{e^{x}\sin x+C}$$
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NCERT Question 17: Evaluate the integral
$$\int \frac{x e^{x}}{(1+x)^{2}}\;dx$$
Solution
Let
$$f(x)=\int \frac{x e^{x}}{(1+x)^{2}}\;dx$$
$$f(x)=\int e^{x}\left(\frac{x}{(1+x)^{2}}\right)\;dx$$
$$f(x)=\int e^{x}\left(\frac{1+x-1}{(1+x)^{2}}\right)\;dx$$
$$f(x)=\int e^{x}\left(\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\right)\;dx$$
$$\text{Let }u=\frac{1}{1+x}$$
$$u’=-\frac{1}{(1+x)^{2}}$$
$$f(x)=\int e^{x}{[u+u’]}\;dx$$
$$\int e^{x}{[u+u’]}\;dx=e^{x}u+C$$
$$f(x)=\int e^{x}\left(\frac{1}{1+x}+\frac{-1}{(1+x)^{2}}\right)\;dx$$
$$f(x)=\frac{e^{x}}{1+x}+C$$
Final Answer
$$\boxed{\frac{e^{x}}{1+x}+C}$$
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NCERT Question 18: Evaluate the integral
$$\displaystyle \int e^{x}\;\frac{1+\sin x}{1+\cos x}\;dx$$
Solution
$$\displaystyle I=\int e^{x}\;\frac{1+\sin x}{1+\cos x}\;dx$$
Use half-angle identities. Note that
$$1+\cos x=2\cos^{2}\frac{x}{2}$$
and
$$1+\sin x=\biggl(\sin\frac{x}{2}+\cos\frac{x}{2}\biggr)^{2}.$$
Therefore
$$\frac{1+\sin x}{1+\cos x}
=\frac{\bigl(\sin\frac{x}{2}+\cos\frac{x}{2}\bigr)^{2}}{2\cos^{2}\frac{x}{2}}.$$
Rewrite the fraction:
$$\frac{1+\sin x}{1+\cos x}
=\tfrac{1}{2}\biggl(\frac{\sin\frac{x}{2}+\cos\frac{x}{2}}{\cos\frac{x}{2}}\biggr)^{2}.$$
Simplify the quotient inside the square:
$$\frac{\sin\frac{x}{2}+\cos\frac{x}{2}}{\cos\frac{x}{2}}
=1+\tan\frac{x}{2}.$$
Hence
$$\frac{1+\sin x}{1+\cos x}
=\tfrac{1}{2}\bigl(1+\tan\frac{x}{2}\bigr)^{2}.$$
Expand using $1+\tan^{2}\theta=\sec^{2}\theta$:
$$\tfrac{1}{2}\bigl(1+\tan\frac{x}{2}\bigr)^{2}
=\tfrac{1}{2}\bigl(\sec^{2}\frac{x}{2}+2\tan\frac{x}{2}\bigr).$$
So the integrand becomes
$$e^{x}\;\frac{1+\sin x}{1+\cos x}
=e^{x}\Bigl(\tfrac{1}{2}\sec^{2}\frac{x}{2}+\tan\frac{x}{2}\Bigr).$$
Let
$$u=\tan\frac{x}{2}.$$
Then
$$u’=\frac{1}{2}\sec^{2}\frac{x}{2}.$$
Thus the integrand is $e^{x}{[u+u’]}$ and
$$I=\int e^{x}{[u+u’]}\;dx.$$
Use the standard result $\displaystyle \int e^{x}{[u’+u]}\;dx=e^{x}u+C$. Therefore
$$I=e^{x}\Bigl(\tfrac{1}{2}\sec^{2}\frac{x}{2}+\tan\frac{x}{2}\Bigr)=e^{x}\tan\frac{x}{2}+C.$$
Final Answer
$$\boxed{\displaystyle \int e^{x}\;\frac{1+\sin x}{1+\cos x}\;dx=e^{x}\tan\frac{x}{2}+C}$$
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NCERT Question 19: Evaluate the integral
$$\displaystyle \int e^{x}\Bigl(\frac{1}{x}-\frac{1}{x^{2}}\Bigr)\;dx$$
Solution
$$\displaystyle I=\int e^{x}\Bigl(\frac{1}{x}-\frac{1}{x^{2}}\Bigr)\;dx$$
Let
$$f(x)=\frac{1}{x}$$
then
$$f'(x)=-\frac{1}{x^{2}}.$$
Using the result
$$\int e^{x}{[f(x)+f'(x)]}\;dx=e^{x}f(x)+C\;$$
we obtain
$$\displaystyle I=\int e^{x}\Bigl(\frac{1}{x}+\left[-\frac{1}{x^{2}}\right]\Bigr)\;dx=e^{x}\;\frac{1}{x}+C.$$
Final Answer
$$\boxed{\displaystyle \int e^{x}\Bigl(\frac{1}{x}-\frac{1}{x^{2}}\Bigr)\;dx=\frac{e^{x}}{x}+C}$$
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NCERT Question 20: Evaluate the integral
$$\displaystyle \int e^{x}\;\frac{x-3}{(x-1)^{3}}\;dx$$
Solution
We begin with
$$I=\int e^{x}\;\frac{x-3}{(x-1)^{3}}\;dx.$$
Rewrite the numerator:
$$(x-3)=(x-1)-2.$$
Thus,
$$I=\int e^{x}\Bigl(\frac{1}{(x-1)^{2}}-\frac{2}{(x-1)^{3}}\Bigr)\;dx.$$
Let
$$f(x)=\frac{1}{(x-1)^{2}}.$$
Then
$$f'(x)=-\frac{2}{(x-1)^{3}}.$$
Using the standard result
$$\int e^{x}{[f(x)+f'(x)]}\;dx=e^{x}f(x)+C\;$$
$$I=\int e^{x}\Bigl(\frac{1}{(x-1)^{2}}-\frac{2}{(x-1)^{3}}\Bigr)\;dx=e^{x}\;\frac{1}{(x-1)^{2}}+C $$
we get
$$I=e^{x}\;\frac{1}{(x-1)^{2}}+C.$$
Final Answer
$$\boxed{\displaystyle \int e^{x}\frac{x-3}{(x-1)^{3}}\;dx=\frac{e^{x}}{(x-1)^{2}}+C}$$
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NCERT Question 21: Evaluate the integral
$$\int e^{2x}\sin x\;dx$$
Solution
$$f(x)=\int e^{2x}\sin x\;dx$$
$$f(x)=\sin x\int e^{2x}\;dx-\int\frac{d(\sin x)}{dx}\left(\int e^{2x}\;dx\right)\;dx$$
$$\int e^{2x}\;dx=\frac{e^{2x}}{2}$$
$$f(x)=\sin x\cdot\frac{e^{2x}}{2}-\int \cos x\cdot\frac{e^{2x}}{2}\;dx$$
$$f(x)=\frac{e^{2x}}{2}\sin x-\frac{1}{2}\int e^{2x}\cos x\;dx$$
Now evaluate the remaining integral by parts.
$$\int e^{2x}\cos x\;dx
=\cos x\int e^{2x}\;dx-\int\frac{d(\cos x)}{dx}\left(\int e^{2x}\;dx\right)\;dx$$
$$\int e^{2x}\cos x\;dx
=\cos x\cdot\frac{e^{2x}}{2}-\int(-\sin x)\cdot\frac{e^{2x}}{2}\;dx$$
$$\int e^{2x}\cos x\;dx
=\frac{e^{2x}}{2}\cos x+\frac{1}{2}\int e^{2x}\sin x\;dx$$
Substitute this into the expression for $f(x)$.
$$f(x)=\frac{e^{2x}}{2}\sin x-\frac{1}{2}\left(\frac{e^{2x}}{2}\cos x+\frac{1}{2}\int e^{2x}\sin x\;dx\right)$$
$$f(x)=\frac{e^{2x}}{2}\sin x-\frac{e^{2x}}{4}\cos x-\frac{1}{4}\int e^{2x}\sin x\;dx$$
Bring the integral term to the left side.
$$f(x)+\frac{1}{4}f(x)=\frac{e^{2x}}{2}\sin x-\frac{e^{2x}}{4}\cos x$$
$$\frac{5}{4}f(x)=\frac{e^{2x}}{2}\sin x-\frac{e^{2x}}{4}\cos x$$
Multiply both sides by $\tfrac{4}{5}$.
$$f(x)=\frac{4}{5}\left(\frac{e^{2x}}{2}\sin x-\frac{e^{2x}}{4}\cos x\right)$$
Simplify the coefficients.
$$f(x)=\frac{e^{2x}}{5}\bigl(2\sin x-\cos x\bigr)+C$$
Final Answer
$$\boxed{\displaystyle \int e^{2x}\sin x\;dx=\frac{e^{2x}}{5}\bigl(2\sin x-\cos x\bigr)+C}$$
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NCERT Question 22: Evaluate the integral
$$\displaystyle \int \sin^{-1}\left(\frac{2x}{1+x^{2}}\right)\;dx$$
Solution
Let
$$I=\int \sin^{-1}\left(\frac{2x}{1+x^{2}}\right)\;dx$$
Substitute $x=\tan\theta$ so that
$$dx=\sec^{2}\theta\;d\theta$$
and
$$\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)
=\sin^{-1}\left(\frac{2\tan\theta}{1+\tan^{2}\theta}\right)
=\sin^{-1}(\sin 2\theta).$$
(Assume $\lvert x\rvert\le 1$ so $2\theta\in\bigl(-\tfrac{\pi}{2}\;\tfrac{\pi}{2}\bigr)$ and $\sin^{-1}(\sin 2\theta)=2\theta$.)
Hence the integral becomes
$$I=\int 2\theta\sec^{2}\theta\;d\theta.$$
Integrate by parts. Take
$$u=2\theta\quad\text{and}\quad dv=\sec^{2}\theta\;d\theta.$$
Then
$$du=2\;d\theta\quad\text{and}\quad v=\tan\theta.$$
So
$$I=uv-\int v\;du
=2\theta\tan\theta-2\int \tan\theta\;d\theta.$$
Evaluate the remaining integral:
$$\int \tan\theta\;d\theta=-\ln\lvert\cos\theta\rvert.$$
Thus
$$I=2\theta\tan\theta+2\ln\lvert\cos\theta\rvert + C.$$
Return to $x$ using $\theta=\tan^{-1}x$, $\tan\theta=x$ and
$$\ln\lvert\cos\theta\rvert=\ln\left(\frac{1}{\sqrt{1+x^{2}}}\right)=-\tfrac{1}{2}\ln(1+x^{2}).$$
Substitute these back:
$$I=2x\tan^{-1}x- \ln(1+x^{2}) + C.$$
Final Answer
$$\boxed{\displaystyle \int \sin^{-1}\left(\frac{2x}{1+x^{2}}\right)\;dx
=2x\tan^{-1}x- \ln(1+x^{2}) + C}$$
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NCERT Question 23: Choose the correct answer by evaluate the integral
$$\int x^{2} e^{x^{3}}\;dx$$
Options
(A) $\frac{1}{3}e^{x^{3}}+C$
(B) $\frac{1}{3}e^{X^{3}}+C$
(C) $\frac{1}{2}e^{X^{3}}+C$
(D) $\frac{1}{3}e^{X^{3}}+C$
Solution
Let
$$I=\int x^{2} e^{x^{3}}\;dx\;$$
Use the substitution
$$t=x^{3},\qquad dt=3x^{2}\;dx$$
which gives
$$x^{2}\;dx=\frac{dt}{3}.$$
Now the integral becomes
$$I=\int e^{t}\;\frac{dt}{3}
=\frac{1}{3}\int e^{t}\;dt.$$
Thus,
$$I=\frac{1}{3}e^{t}+C
=\frac{1}{3}e^{x^{3}}+C.$$
Correct Answer
(A) $\dfrac{1}{3}e^{x^{3}} + C $
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NCERT Question 24: Choose the correct answer by evaluate the integral
$$\int e^{x}\;\sec x\;(1+\tan x)\;dx$$
Options
(A) $e^{x}\cos x + C$
(B) $e^{x}\sec x + C$
(C) $e^{x}\sin x + C$
(D) $e^{x}\tan x + C$
Solution
Let
$$I=\int e^{x}\;\sec x\;(1+\tan x)\;dx.$$
Rewrite the integrand:
$$\sec x\;(1+\tan x)=\sec x+\sec x\;\tan x.$$
So
$$I=\int e^{x}(\sec x+\sec x\;\tan x)\;dx.$$
Observe that
$$\frac{d}{dx}(\sec x)=\sec x\;\tan x.$$
Therefore
$$e^{x}(\sec x+\sec x\;\tan x)=\frac{d}{dx}(e^{x}\sec x).$$
Hence
$$I=\int \frac{d}{dx}(e^{x}\sec x)\;dx.$$
Thus
$$I=e^{x}\sec x + C.$$
Correct Answer
(B) $e^{x}\sec x + C $
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