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Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 19

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Hint: You must know about odd and even function.

Given: \int_{-1}^{1} \log \left(\frac{2-x}{2+x}\right){\mathrm{dx}}


\mathrm{I}=\int_{-1}^{1} \log \left(\frac{2-x}{2+x}\right) \mathrm{dx}

Firstly, we check f (x) is odd or even.

Let   f(x)=\log \left(\frac{2-x}{2+x}\right)

\begin{aligned} \mathrm{f}(-\mathrm{x}) &=\log \left(\frac{2+x}{2-x}\right) \\\\ &=\log (2+x)-\log (2-x) \end{aligned}


            \begin{aligned} &=-[\log (2-x)-\log (2+x)] \\\\ &=-{\log \left(\frac{2-x}{2+x}\right)} \\\\ &=-\mathrm{f}(\mathrm{x}) \end{aligned}

\Rightarrow \mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{f}(\mathrm{x}) is an odd function.

\therefore\int_{-1}^{1} \log \left(\frac{2-x}{2+x}\right){\mathrm{dx}}=0                    [\text { If } f(x)=-f(x) \text { then } f(x) \text { is odd.] }

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