Anand Classes presents well-structured NCERT and study material for Integrals Exercise 7.9 of Chapter 7 for Class 12 Mathematics, designed to help students master integration concepts with clarity and confidence. This Set-1 study resource includes step-by-step solutions, conceptual explanations, and exam-oriented guidance aligned with the latest NCERT syllabus. Click the print button to download study material and notes.
NCERT Question 1: Evaluate the integral
$$\int_0^1 \frac{x}{x^2+1}\;dx$$
Solution
$$\int_0^1 \frac{x}{x^2+1}\;dx$$
Let
$$x^2+1=t \quad\Rightarrow\quad 2x\;dx=dt$$
When
$x=0 \Rightarrow t=1$
and when
$x=1 \Rightarrow t=2$
Thus,
$$\int_0^1 \frac{x}{x^2+1}\;dx=\frac12 \int_1^2 \frac{dt}{t}$$
Now,
$$\frac12 \int_1^2 \frac{dt}{t}=\frac12 \left[\ln|t|\right]_1^2$$
So,
$$\frac12[\ln2-\ln1]=\frac12\ln2$$
Final Result
$$\boxed{\frac12\ln2}$$
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NCERT Question 2: Evaluate the integral
$$I=\int_{0}^{\dfrac{\pi}{2}}\sqrt{\sin\phi}\;\cos^{5}\phi\;d\phi$$
Solution
$$I=\int_{0}^{\dfrac{\pi}{2}}\sqrt{\sin\phi}\;\cos^{5}\phi\;d\phi$$
Use the substitution $t=\sin\phi$ so that $ dt=\cos\phi\;d\phi$.
When $\phi=0\Rightarrow t=0$ and when $\phi=\dfrac{\pi}{2}\Rightarrow t=1$.
Rewrite the integrand:
$$\sqrt{\sin\phi}\;\cos^{5}\phi\;d\phi
=t^{\tfrac{1}{2}}\big(1-t^{2}\big)^{2}\;dt$$
So
$$I=\int_{0}^{1}t^{\tfrac{1}{2}}\big(1-2t^{2}+t^{4}\big)\;dt
=\int_{0}^{1}\big(t^{\tfrac{1}{2}}-2t^{\tfrac{5}{2}}+t^{\tfrac{9}{2}}\big)\;dt$$
Integrate termwise:
$$\int_{0}^{1}t^{\tfrac{1}{2}}\;dt=\left[\frac{2}{3}t^{\tfrac{3}{2}}\right]_0^1=\frac{2}{3}$$
$$\int_{0}^{1}-2t^{\tfrac{5}{2}}dt=-2\left[\frac{2}{7}t^{\tfrac{7}{2}}\right]_{0}^{1}=-\frac{4}{7}$$
$$\int_{0}^{1}t^{\tfrac{9}{2}}dt=\left[\frac{2}{11}t^{\tfrac{11}{2}}\right]_{0}^{1}=\frac{2}{11}$$
Add the results:
$$I=\frac{2}{3}-\frac{4}{7}+\frac{2}{11}$$
Final Result
$$\boxed{I=\frac{64}{231}}$$
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NCERT Question 3: Evaluate the integral
$$I=\int_{0}^{1}\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)\;dx$$
Solution
Substitution. Put $x=\tan\theta$, hence $dx=\sec^{2}\theta\;d\theta$.
When $x=0\Rightarrow\theta=0$, and when $x=1\Rightarrow\theta=\dfrac{\pi}{4}$.
Note that
$$\frac{2x}{1+x^{2}}=\frac{2\tan\theta}{1+\tan^{2}\theta}=2\sin\theta\cos\theta=\sin2\theta,$$
so the integrand becomes $\sin^{-1}(\sin2\theta)=2\theta$ for $0\le\theta\le\dfrac{\pi}{4}$.
Therefore
$$I=\int_{0}^{\dfrac{\pi}{4}}2\theta\sec^{2}\theta\;d\theta.$$
Integrate by parts. Let $u=2\theta$ and $dv=\sec^{2}\theta d\theta$. Then $du=2d\theta$ and $v=\tan\theta$. Thus
$$
I=\big[2\theta\tan\theta\big]_{0}^{\dfrac{\pi}{4}}-2\int_{0}^{\dfrac{\pi}{4}}\tan\theta d\theta
=2\Big[\theta\tan\theta+\ln\big|\cos\theta\big|\Big]_{0}^{\dfrac{\pi}{4}}.
$$
Evaluate the boundary terms. At $\theta=\dfrac{\pi}{4}$ we have $\tan\theta=1$ and $\cos\theta={\dfrac{1}{\sqrt{2}}}$, so
$$
\theta\tan\theta+\ln\big|\cos\theta\big|=\frac{\pi}{4}-\frac{1}{2}\ln 2,
$$
and at $\theta=0$ the expression is $0$. Hence
$$
I=2\Big(\frac{\pi}{4}-\frac{1}{2}\ln 2\Big)=\frac{\pi}{2}-\ln 2.
$$
Final Result
$$\boxed{I=\frac{\pi}{2}-\ln 2}$$
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NCERT Question 4: Evaluate the integral
$$I=\int_{0}^{2}x\sqrt{x+2}\;dx$$
Solution
Let
$$t=x+2 \quad\Rightarrow\quad x=t-2 \quad\text{and}\quad dx=dt.$$
When $x=0 \Rightarrow t=2$ and when $x=2 \Rightarrow t=4$.
So,
$$
I=\int_{2}^{4}(t-2)\sqrt{t}\;dt
$$
$$
=\int_{2}^{4}\left(t^{3/2}-2t^{1/2}\right)\;dt
$$
Integrating term by term:
$$
I=\left(\frac{2}{5}t^{5/2}-\frac{4}{3}t^{3/2}\right)\Bigg|_{2}^{4}
$$
Evaluate at $t=4$:
$$
\frac{2}{5}(4)^{5/2}-\frac{4}{3}(4)^{3/2}
=\frac{2}{5}(32)-\frac{4}{3}(8)
=\frac{64}{5}-\frac{32}{3}
$$
Evaluate at $t=2$:
$$
\frac{2}{5}(2)^{5/2}-\frac{4}{3}(2)^{3/2}
=\frac{2}{5}(4\sqrt{2})-\frac{4}{3}(2\sqrt{2})
=\frac{8\sqrt{2}}{5}-\frac{8\sqrt{2}}{3}
$$
Subtracting:
$$
I=\left(\frac{64}{5}-\frac{32}{3}\right)-\left(\frac{8\sqrt{2}}{5}-\frac{8\sqrt{2}}{3}\right)
$$
Common denominators:
$$
I=\frac{32}{15}+\frac{16\sqrt{2}}{15}
$$
So,
Final Result
$$\boxed{\frac{32+16\sqrt{2}}{15}}$$
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NCERT Question 5: Evaluate the integral
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\sin x}{1+\cos^{2}x}\;dx$$
Solution
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\sin x}{1+\cos^{2}x}\;dx$$
Use the substitution $t=\cos x\Rightarrow dt=-\sin x\;dx$.
When $x=0\Rightarrow t=1$, and when $x=\dfrac{\pi}{2}\Rightarrow t=0$.
Therefore
$$I=-\int_{1}^{0}\frac{dt}{1+t^{2}}=\int_{0}^{1}\frac{dt}{1+t^{2}}.$$
Evaluate the integral. An antiderivative is $\tan^{-1}t$, so
$$I=\left[\tan^{-1}t\right]_{0}^{1}=\tan^{-1}1-\tan^{-1}0=\frac{\pi}{4}-0=\frac{\pi}{4}.$$
Final Result
$$\boxed{I=\frac{\pi}{4}}$$
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NCERT Question 6: Evaluate the integral
$$I=\int_{0}^{2}\frac{dx}{x+4-x^{2}}$$
Solution
$$I=\int_{0}^{2}\frac{dx}{x+4-x^{2}}$$
Rewrite the denominator and complete the square.
$$x+4-x^{2}=-(x^{2}-x-4)=\frac{17}{4}-\bigg(x-\frac{1}{2}\bigg)^{2}.$$
Change variable $u=x-\tfrac{1}{2}$ so $du=dx$. The limits become $u(0)=-\tfrac{1}{2}$ and $u(2)=\tfrac{3}{2}$. Thus
$$I=\int_{-\tfrac{1}{2}}^{\tfrac{3}{2}}\frac{du}{\frac{17}{4}-u^{2}}.$$
Use the standard integral $\displaystyle\int\frac{du}{a^{2}-u^{2}}=\frac{1}{2a}\ln\left|\frac{a+u}{a-u}\right|$ with $a=\dfrac{\sqrt{17}}{2}$. Therefore
$$I=\frac{1}{2a}\left[\ln\left|\frac{a+u}{a-u}\right|\right]_{u=-\tfrac{1}{2}}^{u=\tfrac{3}{2}}
=\frac{1}{\sqrt{17}}\left[\ln\frac{a+\tfrac{3}{2}}{a-\tfrac{3}{2}}-\ln\frac{a-\tfrac{1}{2}}{a+\tfrac{1}{2}}\right].$$
Substitute $a=\dfrac{\sqrt{17}}{2}$ so $a\pm\tfrac{3}{2}=\dfrac{\sqrt{17}\pm3}{2}$ and $a\pm\tfrac{1}{2}=\dfrac{\sqrt{17}\pm1}{2}$. The factors of $1/2$ cancel inside the ratios, giving
$$I=\frac{1}{\sqrt{17}}\ln\left(\frac{\sqrt{17}+3}{\sqrt{17}-3}\cdot\frac{\sqrt{17}+1}{\sqrt{17}-1}\right).$$
$$I=\frac{1}{\sqrt{17}}\ln\left(\frac{5\sqrt{17}+21}{4}\right).$$
Final Result
$$\boxed{I=\frac{1}{\sqrt{17}}\ln\left(\frac{5\sqrt{17}+21}{4}\right)}$$
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NCERT Question 7: Evaluate the integral
$$I=\int_{-1}^{1}\frac{dx}{x^{2}+2x+5}$$
Solution
Complete the square in the denominator.
$$x^{2}+2x+5=(x+1)^{2}+4.$$
Use the substitution $t=x+1\Rightarrow dt=dx$. When $x=-1\Rightarrow t=0$ and when $x=1\Rightarrow t=2$. The integral becomes
$$I=\int_{0}^{2}\frac{dt}{t^{2}+4}.$$
Use the standard antiderivative $\displaystyle\int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{u}{a}$. Here $a=2$, so
$$I=\left[\frac{1}{2}\tan^{-1}\frac{t}{2}\right]_{0}^{2}
=\frac{1}{2}\big(\tan^{-1}1-\tan^{-1}0\big)
=\frac{1}{2}\cdot\frac{\pi}{4}=\frac{\pi}{8}.$$
Final Result
$$\boxed{I=\frac{\pi}{8}}$$
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NCERT Question 8: Evaluate the integral
$$I=\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2x^{2}}\right)e^{2x}\;dx$$
Solution
$$I=\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2x^{2}}\right)e^{2x}\;dx$$
Use the substitution $t=2x\Rightarrow dt=2dx\Rightarrow dx=\dfrac{dt}{2}$.
When $x=1\Rightarrow t=2$ and when $x=2\Rightarrow t=4$.
Rewrite the integrand with $x=\dfrac{t}{2}$:
$$\frac{1}{x}-\frac{1}{2x^{2}}=\frac{2}{t}-\frac{2}{t^{2}}=2\Big(\frac{1}{t}-\frac{1}{t^{2}}\Big).$$
Hence
$$I=\int_{2}^{4}\Big(\frac{1}{t}-\frac{1}{t^{2}}\Big)e^{t}\;dt$$
Recognise the pattern. Let $f(t)=\dfrac{1}{t}$ so $f'(t)=-\dfrac{1}{t^{2}}$. Then $f(t)+f'(t)=\dfrac{1}{t}-\dfrac{1}{t^{2}}$, and
$$\frac{d}{dt}\big(e^{t}f(t)\big)=e^{t}f(t)+e^{t}f'(t)=e^{t}\Big(\frac{1}{t}-\frac{1}{t^{2}}\Big).$$
Therefore an antiderivative is $e^{t}\dfrac{1}{t}$ and
$$I=\left[e^{t}\frac{1}{t}\right]_{2}^{4}=\frac{e^{4}}{4}-\frac{e^{2}}{2}.$$
Final Result
$$\boxed{I=\frac{e^{4}}{4}-\frac{e^{2}}{2}}$$
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NCERT Question 9 : Evaluate the integral
$$\int_{\frac{1}{3}}^{1}\frac{(x – x^{3})^{1/3}}{x^{4}}\;dx$$
Choose the Correct Answer
[A] 6
[B] 0
[C] 3
[D] 4
Solution
Let
$$I=\int_{\frac{1}{3}}^{1}\frac{(x – x^{3})^{1/3}}{x^{4}}\;dx.$$
Substitute
$$x=\frac{1}{y}\;\qquad dx=-\frac{1}{y^{2}}dy.$$
Limits:
- When $x=\frac{1}{3}$, $y=3$
- When $x=1$, $y=1$
Thus,
$$I=\int_{3}^{1}\left(\frac{1}{y}-\frac{1}{y^{3}}\right)^{1/3} y^{4}\left(-\frac{1}{y^{2}}\right)dy.$$
Simplify:
$$\left(\frac{1}{y}-\frac{1}{y^{3}}\right)^{1/3}=\frac{(y^{2}-1)^{1/3}}{y}\;$$
so the integrand becomes:
$$-y(y^{2}-1)^{1/3}.$$
Reverse limits:
$$I=\int_{1}^{3}y(y^{2}-1)^{1/3}\;dy.$$
Substitute:
$$t=y^{2}-1\;\qquad dt=2y\;dy\qquad \Rightarrow\qquad y\;dy=\frac{dt}{2}.$$
Limits:
- $y=1\Rightarrow t=0$
- $y=3\Rightarrow t=8$
So,
$$I=\frac{1}{2}\int_{0}^{8}t^{1/3}\;dt
=\frac{1}{2}\left[\frac{t^{4/3}}{4/3}\right]_{0}^{8}
=\frac{1}{2}\cdot \frac{3}{4}\cdot 16=6.$$
Final Result
$$\boxed{6}$$
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NCERT Question 10 : If
$$f(x)=\int_{0}^{x} t\sin t\;dt$$
then f'(x) is:
[A] $\cos x + x\sin x $
[B] $x\sin x$
[C] $ x\cos x $
[D] $\sin x + x\cos x $
Solution :
We have
$$f(x)=\int_{0}^{x} t\sin t\;dt.$$
This is an integral with a variable upper limit, so we can use the Leibniz Rule / Fundamental Theorem of Calculus:
Important: If
$$F(x)=\int_{a}^{x} g(t)\;dt,$$
then
$$F'(x)=g(x).$$
Here $g(t) = t\sin t$
So,
$$f'(x)=x\sin x.$$
Correct Answer: [B] $x\sin x$
Final Result
$$\boxed{x\sin x}$$
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