Exercise 7.9 (Revised) - Chapter 7 - Integrals - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions: Chapter 7 Integrals | Free PDF Download

Ex 7.9 Question 1.

$\int_0^1 \frac{x}{x^2+1} d x$

Answer.

Let $\mathrm{I}=\int_0^1 \frac{x}{x^2+1} d x$
$=\frac{1}{2} \int_0^1 \frac{2 x}{x^2+1} d x$
Putting $x^2+1=t$
$
\begin{aligned}
& \Rightarrow 2 x=\frac{d t}{d x} \\
& \Rightarrow 2 x d x=d t
\end{aligned}
$

To change the limits of integration from $x$ to $t$
when $\mathrm{x}=0, \mathrm{t}=\mathrm{x}^2+1=0+1=1$
when $\mathrm{x}=1, \mathrm{t}=\mathrm{x}^2+1=1+1=2$
$\therefore$ From eq. (i),
$
\begin{aligned}
& =\frac{1}{2} \int_1^2 \frac{d t}{t} \\
& =\frac{1}{2}(\log |t|)_1^2
\end{aligned}
$

$\begin{aligned}
& =\frac{1}{2}(\log |2|-\log |1|) \\
& =\frac{1}{2}(\log 2-\log 1) \\
& =\frac{1}{2}(\log 2-0) \\
& =\frac{1}{2} \log 2 \text { Ans. }
\end{aligned}$

Ex 7.9 Question 2.

$\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^5 \phi d \phi$

Answer.

Let $I=\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^5 \phi d \phi$
Putting $\sin \phi=t$
$\Rightarrow \cos \phi=\frac{d t}{d \phi}$
$\Rightarrow \cos \phi d \phi=d t$
To change the limits of integration from $\phi$ to $t$
When $\phi=0, t=\sin \phi=\sin 0^{\circ}=0$
When $\phi=\frac{\pi}{2}, t=\sin \phi=\sin \frac{\pi}{2}=1$
$\therefore$ From eq. (i),

$\begin{aligned}
& \mathrm{I}=\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^4 \phi \cos \phi d \phi \\
& =\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi}\left(\cos ^2 \phi\right)^2 \cos \phi d \phi \\
& =\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi}\left(1-\sin ^2 \phi\right)^2 \cos \phi d \phi \\
& =\int_0^1 \sqrt{t}\left(1-t^2\right)^2 d t \\
& =\int_0^1 t^{\frac{1}{2}}\left(1+t^4-2 t^2\right) d t \\
& =\int_0^1\left(t^{\frac{1}{2}}+t^{\frac{1}{2}+4}-2 t^{\frac{1}{2}+2}\right) d t \\
& =\int_0^1\left(t^{\frac{1}{2}}+t^{\frac{9}{2}}-2 t^{\frac{5}{2}}\right) d t \\
& =\int_0^1 t^{\frac{1}{2}} d t+\int_0^1 t^{\frac{9}{2}} d t-2 \int_0^1 t^{\frac{5}{2}} d t \\
& =\frac{2}{3}\left(t^{\frac{3}{2}}\right)_0^1+\frac{2}{11}\left(t^{\frac{11}{2}}\right)_0^1-\frac{4}{7}\left(t^{\frac{7}{2}}\right)_0^1 \\
& =\frac{2}{3}(1-0)+\frac{2}{11}(1-0)-\frac{4}{7}(1-0) \\
&
\end{aligned}$

$=\frac{2}{3}+\frac{2}{11}-\frac{4}{7}$

$\begin{aligned}
& =\frac{154+42+132}{231} \\
& =\frac{64}{231} \text { Ans. }
\end{aligned}$

Ex 7.9 Question 3.

$\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$

Answer.

Let $\mathrm{I}=\int_0^1 \sin ^{-1}\left(\frac{2 x}{1+x^2} d x\right)$
Putting $x=\tan \theta$
$
\begin{aligned}
& \Rightarrow \frac{d x}{d \theta} \sec ^2 \theta \\
& \Rightarrow d x=\sec ^2 \theta d \theta
\end{aligned}
$

Limits of integration, when $x=0, \tan \theta=0=\tan 0^{\circ} \Rightarrow \theta=0$
when $x=1, \tan \theta=1=\tan \frac{\pi}{4}$
$
\Rightarrow \theta=\frac{\pi}{4}
$
$\therefore$ From eq. (i),
$
\begin{aligned}
& \mathrm{I}=\int_0^{\frac{\pi}{4}}\left(\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\right) \sec ^2 \theta d \theta \\
& =\int_0^{\frac{\pi}{4}}\left(\sin ^{-1}(\sin 2 \theta)\right) \sec ^2 \theta d \theta
\end{aligned}
$

$
\begin{aligned}
& =\int_0^{\frac{\pi}{4}} 2 \theta \sec ^2 \theta d \theta \\
& =2 \int_0^{\frac{\pi}{4}} \theta \sec ^2 \theta d \theta
\end{aligned}
$
[Applying Product Rule]
$
\begin{aligned}
& =2\left[(\theta \cdot \tan \theta)_0^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}} 1 \cdot \tan \theta d \theta\right] \\
& =2\left[\frac{\pi}{4} \tan \frac{\pi}{4}-0-\int_0^{\frac{\pi}{4}} \tan \theta d \theta\right] \\
& =2\left[\frac{\pi}{4}-(\log \sec \theta)_0^{\frac{\pi}{4}}\right] \\
& =2\left[\frac{\pi}{4}-\left(\log \sec \frac{\pi}{4}-\log \sec 0^{\circ}\right)\right] \\
& =2\left[\frac{\pi}{4}-(\log \sqrt{2}-\log 1)\right] \\
& =\frac{\pi}{2}-2 \log 2^{\frac{1}{2}} \\
& =\frac{\pi}{2}-2 \cdot \frac{1}{2} \log 2
\end{aligned}
$

$=\frac{\pi}{2}-\log 2 \text { Ans. }$

Ex 7.9 Question 4.

$\int_0^2 x \sqrt{x+2} d x$

Answer.

Let $\mathrm{I}=\int_0^2 x \sqrt{x+2} d x$
Putting $\sqrt{x+2}=t$
$
\begin{aligned}
& \Rightarrow x+2=t^2 \\
& \Rightarrow \frac{d x}{d t}=2 t \\
& \Rightarrow d x=2 t d t
\end{aligned}
$

Limits of integration when $x=0, t=\sqrt{2}$ and when $x=2, t=\sqrt{4}=2$
$\therefore$ From eq. (i),
$
\begin{aligned}
\mathrm{I} & =\int_{\sqrt{2}}^2\left(t^2-2\right) t \cdot 2 t d t \\
& =2 \int_{\sqrt{2}}^2 t^2\left(t^2-2\right) d t \\
& =2 \int_{\sqrt{2}}^2\left(t^4-2 t^2\right) d t
\end{aligned}
$

$\begin{aligned}
& =2\left[\left(\frac{t^5}{5}\right)_{\sqrt{2}}^2-2\left(\frac{t^3}{3}\right)_{\sqrt{2}}^2\right] \\
& =2\left[\frac{1}{5}\left(2^5-(\sqrt{2})^5\right)-\frac{2}{3}\left(2^3-(\sqrt{2})^3\right)\right]
\end{aligned}$

$\begin{aligned}
& =2\left[\frac{1}{5}(32-4 \sqrt{2})-\frac{2}{3}(8-2 \sqrt{2})\right] \\
& =2\left[\frac{32}{5}-\frac{4 \sqrt{2}}{5}-\frac{16}{3}+\frac{4 \sqrt{2}}{3}\right] \\
& =2\left[\frac{96-12 \sqrt{2}-80+20 \sqrt{2}}{15}\right] \\
& =\frac{2}{15}(16+8 \sqrt{2}) \\
& =\frac{16 \sqrt{2}}{15}(\sqrt{2}+1) \text { Ans. }
\end{aligned}$

Ex 7.9 Question 5.

$\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x} d x$

Answer.

Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x} d x$
$=-\int_0^{\frac{\pi}{2}} \frac{-\sin x}{1+\cos ^2 x} d x$
Putting $\cos x=t$
$
\begin{aligned}
& \Rightarrow-\sin x=\frac{d t}{d x} \\
& \Rightarrow-\sin x d x=d t
\end{aligned}
$

Limits of integration when $x=0, t=\cos 0^{\circ}=1$ and when $x=\frac{\pi}{2}, t=\cos \frac{\pi}{2}=0$

$\begin{aligned}
& \therefore \text { From eq. (i), I }=-\int_1^0 \frac{d t}{1+t^2} \\
& =-\int_1^0 \frac{1}{1+t^2} d t \\
& =-\left(\tan ^{-1} t\right)_0^1 \\
& =-\left(\tan ^{-1} 0-\tan ^{-1} 1\right) \\
& =-\left(0-\frac{\pi}{4}\right)=\frac{\pi}{4} \text { Ans. }
\end{aligned}$

Ex 7.9 Question 6.

$\int_0^2 \frac{d x}{x+4-x^2} d x$

Answer.

$\int_0^2 \frac{d x}{x+4-x^2}$
$
=\int_0^2 \frac{d x}{-x^2+x+4}
$
$=\int_0^2 \frac{d x}{-\left(x^2-x-4\right)}$
$=\int_0^2 \frac{d x}{-\left(x^2-x+\frac{1}{4}-\frac{1}{4}-4\right)}$
$
=\int_0^2 \frac{d x}{-\left[\left(x-\frac{1}{2}\right)^2-\frac{17}{4}\right]}
$

$\begin{aligned}
& =\int_0^2 \frac{d x}{\left(\frac{\sqrt{17}}{2}\right)^2-\left(x-\frac{1}{2}\right)^2} \\
& =\frac{1}{2 \times \frac{\sqrt{17}}{2}}\left[\log \left|\frac{\frac{\sqrt{17}}{2}+\left(x-\frac{1}{2}\right)}{\frac{\sqrt{17}}{2}-\left(x-\frac{1}{2}\right)}\right|_0^2\right. \\
& {\left[\because \int \frac{1}{a^2-x^2} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\right]} \\
& =\frac{1}{\sqrt{17}}\left[\log \left|\frac{\sqrt{17}+2 x-1}{\sqrt{17}-2 x+1}\right|\right]_0^2
\end{aligned}$

$\begin{aligned}
& =\frac{1}{\sqrt{17}}\left[\log \left|\frac{\sqrt{17}+3}{\sqrt{17}-3}\right|-\log \left|\frac{\sqrt{17}-1}{\sqrt{17}+1}\right|\right] \\
& =\frac{1}{\sqrt{17}} \log \left(\frac{\sqrt{17}+3}{\sqrt{17}-3} \times \frac{\sqrt{17}+1}{\sqrt{17}-1}\right) \\
& =\frac{1}{\sqrt{17}} \log \left(\frac{20+4 \sqrt{17}}{20-4 \sqrt{17}}\right) \\
& =\frac{1}{\sqrt{17}} \log \frac{4(5+\sqrt{17})}{4(5-\sqrt{17})} \\
& =\frac{1}{\sqrt{17}} \log \frac{(5+\sqrt{17})}{(5-\sqrt{17})}
\end{aligned}$

$
\begin{aligned}
& =\frac{1}{\sqrt{17}} \log \left(\frac{5+\sqrt{17}}{5-\sqrt{17}} \times \frac{5+\sqrt{17}}{5+\sqrt{17}}\right) \\
& =\frac{1}{\sqrt{17}} \log \frac{(5+\sqrt{17})^2}{25-17} \\
& =\frac{1}{\sqrt{17}} \log \frac{42+10 \sqrt{17}}{8} \\
& =\frac{1}{\sqrt{17}} \log \frac{21+5 \sqrt{17}}{4} \text { Ans. }
\end{aligned}
$
Ex 7.9 Question 7.

$\int_{-1}^1 \frac{d x}{x^2+2 x+5}$

Answer.

Let $\mathrm{I}=\int_{-1}^1 \frac{d x}{x^2+2 x+5}$
$
\begin{aligned}
& =\int_{-1}^1 \frac{d x}{x^2+2 x+1+4} \\
& =\int_{-1}^1 \frac{d x}{(x+1)^2+2^2} \ldots \ldots \ldots \text { (i) }
\end{aligned}
$

Putting $x+1=t$
$
\begin{aligned}
& \Rightarrow 1=\frac{d t}{d x} \\
& \Rightarrow d x=d t
\end{aligned}
$

$\text { Limits of integration when } x=-1, t=-1+1=0 \text { and when } x=1, t=1+1=2$

$\begin{aligned}
& \therefore \text { From eq. (i), } \mathrm{I}=\int_0^2 \frac{1}{t^2+2^2} d t \\
& =\frac{1}{2}\left(\tan ^{-1} \frac{t}{2}\right)_0^2 \\
& {\left[\because \int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right]} \\
& =\frac{1}{2}\left[\tan ^{-1} \frac{2}{2}-\tan ^{-1} \frac{0}{2}\right] \\
& =\frac{1}{2}\left[\tan ^{-1} 1-\tan ^{-1} 0\right] \\
& =\frac{1}{2}\left(\frac{\pi}{4}-0\right)=\frac{\pi}{8} \text { Ans. }
\end{aligned}$

Ex 7.9 Question 8

$\int_1^2\left(\frac{1}{x}-\frac{1}{2 x^2}\right) e^{2 x} d x$
Answer.

Let $\mathrm{I}=\int_1^2\left(\frac{1}{x}-\frac{1}{2 x^2}\right) e^{2 x} d x$
Putting $2 x=t$
$
\begin{aligned}
& \Rightarrow 2=\frac{d t}{d x} \\
& \Rightarrow 2 d x=d t \\
& \Rightarrow d x=\frac{d t}{2}
\end{aligned}
$

Limits of integration when $x=1, t=2 \times 1=2$ and when $x=2, t=2 \times 2=4$
$\therefore$ From eq. (i),

$\begin{aligned}
\mathrm{I} & =\int_2^4\left(\frac{1}{\frac{t}{2}}-\frac{1}{2\left(\frac{t}{2}\right)^2}\right) e^t \frac{d t}{2} \\
& =\int_2^4\left(\frac{2}{t}-\frac{2}{t^2}\right) e^t \frac{d t}{2} \\
& =\int_2^4 \frac{1}{2} .2\left(\frac{1}{t}-\frac{1}{t^2}\right) e^t d t \\
& =\int_2^4\left(\frac{1}{t}-\frac{1}{t^2}\right) e^t d t \\
& =\int_2^4\left\{f(t)+f^{\prime}(t)\right\} e^t d t \\
& =\left\{e^t f(t)\right\}_2^4 \\
& =\left(\frac{e^t}{t}\right)_2^4 \\
& =\frac{e^4}{4}-\frac{e^2}{2} \\
& =\frac{e^4-2 e^2}{4} \\
& =\frac{e^2\left(e^2-2\right)}{4} \text { Ans. }
\end{aligned}$

Ex 7.9 Question 9.

The value of the integral $\int_{\frac{1}{3}}^1 \frac{\left(x-x^3\right)^{\frac{1}{3}}}{x^4} d x$ is:
(A) 6
(B) 0
(C) 3
(D) 4

Answer.

Let $\mathrm{I}=\int_{\frac{1}{3}}^1 \frac{\left(x-x^3\right)^{\frac{1}{3}}}{x^4} d x$
$
\begin{aligned}
& =\int_{\frac{1}{3}}^1 \frac{\left.x^3\left(\frac{x}{x^3}-1\right)\right]^{\frac{1}{3}}}{x^4} d x \\
& =\int_{\frac{1}{3}}^1 \frac{x\left(x^{-2}-1\right)^{\frac{1}{3}}}{x^4} d x \\
& =\int_{\frac{1}{3}}^1\left(x^{-2}-1\right)^{\frac{1}{3}} x^{-3} d x \\
& =\frac{-1}{2} \int_{\frac{1}{3}}^1\left(x^{-2}-1\right)^{\frac{1}{3}}\left(-2 x^{-3}\right) d x
\end{aligned}
$

$\begin{aligned}
& \text { Putting } x^{-2}-1=t \\
& \Rightarrow-2 x^{-3}=\frac{d t}{d x}
\end{aligned}$

$
\Rightarrow-2 x^{-3} d x=d t
$

Limits of integration when
$
\begin{aligned}
& x=\frac{1}{3}, t=\left(\frac{1}{3}\right)^{-2}-1=3^2-1=8 \text { and when } x=1, t=(1)^{-2}-1=1-1=0 \\
& \therefore \text { From eq. (i), } \\
& \mathrm{I}=\frac{-1}{2} \int_8^0 t^{\frac{1}{3}} d t \\
& =\frac{-1}{2}\left[\frac{t^{\frac{4}{3}}}{\frac{4}{3}}\right)_8^0 \\
& =\frac{-1}{2} \cdot \frac{3}{4}\left[0-8^{\frac{4}{3}}\right] \\
& =\frac{-3}{8}\left[-(2)^{3 \times \frac{4}{3}}\right] \\
& =\frac{-3}{8}\left(-2^4\right) \\
& =\frac{3}{8} \times 16=6
\end{aligned}
$

Therefore, option (A) is correct.

Ex 7.9 Question 10.

If $f(x)=\int_0^x t \sin t d t$, then $f^{\prime}(x)$ is:
(A) $\cos x+x \sin x$
(B) $x \sin x$

(C) $x \cos x$
(D) $\sin x+x \cos x$

Answer.

$f(x)=\int_0^x t \sin t d t$
$
=\{t(-\cos t)\}_0^x-\int_0^x 1(-\cos t) d t
$
[Applying Product Rule]
$
\begin{aligned}
& =-x \cos x-0+\int_0^x \cos t d t \\
& =-x \cos x+(\sin t)_0^x \\
& =-x \cos x+\sin x-\sin 0^{\circ} \\
& =-x \cos x+\sin x \\
& \therefore f^{\prime}(x)=-\{x(-\sin x)+(\cos x) 1\}+\cos x \\
& =x \sin x-\cos x+\cos x=x \sin x
\end{aligned}
$

Therefore, option (B) is correct.