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  • Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 54.

Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 54.
 

Answers (1)

Answer: a^2\left [ \frac{\pi}{4}-\frac{1}{2} \right ]  

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

Given: \int_{0}^{a} x \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}} d x

Solution:

\int_{0}^{a} x \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}} d x
Put x^2=a^2\cos 2\theta \Rightarrow 2x dx=a^2-\sin2\theta .2d\theta
\Rightarrow dx=-a^2\sin 2\theta x\; d\theta
When x=0 then \theta =\frac{\pi}{4} & when x=a then \theta =0
\begin{aligned} &\therefore \int_{0}^{a} x \sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}} d x=\int_{\frac{\pi}{4}}^{0} \sqrt{\frac{a^{2}(1-\cos 2 \theta)}{a^{2}(1+\cos 2 \theta)}}\left(-a^{2} \sin 2 \theta\right) d x \\ &=-a^{2} \int_{\frac{\pi}{4}}^{0} \sqrt{\frac{a^{2} 2 \sin ^{2} \theta}{a^{2} 2 \cos ^{2} \theta}}(\sin 2 \theta) d x \\ &=-a^{2} \int_{\frac{\pi}{4}}^{0} \frac{\sin \theta}{\cos \theta}(2 \sin \theta \cos \theta) d \theta \mid \end{aligned}

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

 

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