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  • Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 55.

Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 55.
 

Answers (1)

Answer: \pi a

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

Given: \int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx

Solution:

\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}dx
Put x=a\cos 2\theta \Rightarrow dx=-2a\sin 2\theta d\theta
When x=-a then \theta =\frac{\pi}{2}  &  when x=a then \theta =0
\begin{aligned} &\therefore \int_{-a}^{a} \sqrt{\frac{a-x}{a+x}} d x=\int_{\frac{\pi}{2}}^{0} \sqrt{\frac{a-a \cos 2 \theta}{a+a \cos 2 \theta}}(-2 a \sin 2 \theta) d \theta \\ &=-2 a \int_{\frac{\pi}{2}}^{0} \sqrt{\frac{2 \sin ^{2} \theta}{2 \cos ^{2} \theta}} 2 \sin \theta \cos \theta d x \\ &=-2 a \int_{\frac{\pi}{2}}^{0} \frac{\sin \theta}{\cos \theta}(2 \sin \theta \cos \theta) d \theta \\ &=-2 a \int_{\frac{\pi}{2}}^{0}\left(2 \sin ^{2} \theta\right) d \theta \end{aligned}

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

 

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