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

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

Answers (1)

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}

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