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  • Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Very short answer type question 18

Provide solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Very short answer type question 18

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Answer: 0

Hint: You must know the integration rules of trigonometric function with its limits


Given: \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{a-\sin \theta}{a+\sin \theta}\right) d \theta

 

Solution:  I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{a-\sin \theta}{a+\sin \theta}\right) d \theta            .............(i)

Let  f(x)=\log \left(\frac{a-\sin \theta}{a+\sin \theta}\right)

     f(-x)=\log \left(\frac{a-\sin (-\theta)}{a+\sin (-\theta)}\right)

                   =\log \left(\frac{a+\sin \theta}{a-\sin \theta}\right) \quad[\because \sin (-x)=-\sin x]

                   \begin{aligned} &=-\log \left(\frac{a-\sin \theta}{a+\sin \theta}\right) \\ &=-\mathrm{f}(\mathrm{x}) \end{aligned}

Hence, f(x) is an odd function.

 

Since, \int_{-a}^{a} f(x) d x=0 if f(x) is an odd.

 

\therefore \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{a-\sin \theta}{a+\sin \theta}\right) d \theta=0

 

 

Posted by

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