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

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

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Answer: 0

Hint: You must know about odd and even function.

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

Solution:  

\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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