#### Need Solution for R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 14 Maths Textbook Solution.

Answer:$2log2-1$

Hint: Use indefinite formula then put the limit to get the required answer

Given: $\int_{0}^{1}\left ( \frac{1-x}{1+x} \right )dx$

Solution:

$\int_{0}^{1}\left ( \frac{1-x}{1+x} \right )dx$

Put

\begin{aligned} &x=\cos 2 \theta \Rightarrow 2 \theta=\cos ^{-1} x \Rightarrow \theta=\frac{\cos ^{-1} x}{2} \\ &d x=-\sin 2 \theta(2) d \theta \Rightarrow d x=-2 \sin 2 \theta d \theta \end{aligned}

When$x=0$ then$\theta =\frac{\pi }{2 }\times \frac{1}{2}=\frac{\pi }{4}$

When $x=1$ then $\theta =0$

Then

\begin{aligned} &\int_{0}^{1}\left(\frac{1-x}{1+x}\right) d x=\int_{\frac{\pi}{4}}^{0} \frac{1-\cos 2 \theta}{1+\cos 2 \theta}(-2 \sin 2 \theta) d \theta \\ &=-\int_{\frac{\pi}{4}}^{0} \frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta} 2 \sin 2 \theta d \theta \end{aligned}                        $\left[\begin{array}{l} 1+\cos 2 \theta=2 \cos ^{2} \theta \\ 1-\cos 2 \theta=2 \sin ^{2} \theta \end{array}\right]$

\begin{aligned} &=\int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2} \theta}{\cos ^{2} \theta} 2 \sin 2 \theta d \theta \quad\left[\int_{a}^{b} f(x) d x=-\int_{b}^{a} f(x) d x\right] \\ &=\int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2} \theta}{\cos ^{2} \theta} 2.2 \sin \theta \cos \theta d \theta \quad[\sin 2 \theta=2 \sin \theta \cos \theta] \\ &=4 \int_{0}^{\frac{\pi}{4}} \frac{\sin ^{3} \theta}{\cos \theta} d \theta \end{aligned}

Again put $\cos \theta =t\Rightarrow \sin \theta d\theta =dt\Rightarrow d\theta =\frac{dt}{-\sin \theta }$

When $\theta =0$ then $t=2$

And When $\theta =\frac{\pi }{4}$ then $t=\frac{1}{\sqrt{2}}$

Then

\begin{aligned} &\int_{0}^{1} \frac{1-x}{1+x} d x=4 \int_{1}^{\frac{1}{\sqrt{2}}} \frac{\sin ^{3} \theta}{t} \frac{d t}{-\sin \theta}=-4 \int_{1}^{\frac{1}{\sqrt{2}}} \frac{\sin ^{2} \theta}{t} d t \\ &=-4 \int_{2}^{\frac{1}{\sqrt{2}} 1-\cos ^{2} t}{t} d t \end{aligned}

$\left [ \because \sin ^{2}\theta =1-\cos ^{2}\theta \right ]$

\begin{aligned} &=-4 \int_{2}^{\frac{1}{\sqrt{2}} 1-t^{2}}{t} d t \\ &=-4 \int_{1}^{\frac{1}{\sqrt{2}}}\left(\frac{1}{t}-\frac{t^{2}}{t}\right) d t \\ &=-4 \int_{1}^{\frac{1}{\sqrt{2}}}\left(\frac{1}{t}-t\right) d t \\ &=-4 \int_{1}^{\frac{1}{\sqrt{2}} \frac{1}{t} d t+4 \int_{1}^{\frac{1}{\sqrt{2}}} t d t} \\ &=-4[\log |t|]_{1}^{\frac{1}{\sqrt{2}}}+4\left[\frac{t^{1+1}}{1+1}\right]_{1}^{\frac{1}{\sqrt{2}}} \end{aligned}

$\left[\begin{array}{l} \int x^{n} d x=\frac{x^{n+1}}{n+1} \\ \int \frac{1}{x} d x=\log |x| \end{array}\right]$

\begin{aligned} &=-4[\log |t|]_{1}^{\frac{1}{\sqrt{2}}}+4\left[\frac{t^{2}}{2}\right]_{1}^{\frac{1}{\sqrt{2}}} \\ &=-4\left[\log \frac{1}{\sqrt{2}}-\log 1\right]+2\left[\left(\frac{1}{\sqrt{2}}\right)^{2}-1^{2}\right] \\ &=-4 \log \frac{1}{\sqrt{2}}+2\left(\frac{1}{2}-1\right) \end{aligned}                                            $\left [ log\; \; a^{m}=mlog\: a \right ]$

\begin{aligned} &=-4 \log (\sqrt{2})^{-1}+2\left(-\frac{1}{2}\right) \\ &=-4(-\log \sqrt{2})-1 \\ &=4 \log \sqrt{2}-1=2 \log 2^{\frac{1}{2} \times 2}-1 \\ &=2 \log 2-1 \end{aligned}