Please solve RD Sharma class 12 Chapter Definite integrals exercise 19.2 question 37 maths textbook solution.

Answer        : $\frac{\pi}{2}-1$

Hint              : use indefinite integral formula and the limits to solve this integral

Given           :$\int_{0}^1{}\sqrt{\frac{1-x}{1+x}}dx$

Solution       : $\int_{0}^1{}\sqrt{\frac{1-x}{1+x}}dx$
put $x=\cos 2\theta \Rightarrow dx= -2\sin 2\theta d\theta$
when $x=0 \; \text{then} \; \; \theta =\frac{\pi}{4} \text{ and when}\; x=1 then \theta=0$
therefore ,

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

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