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  • Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 35

Explain solution RD Sharma class 12 Chapter 19 Definite Integrals Exercise Revision Exercise question 35

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Answer:  0

Hint: When the value of f\left ( x \right )  is odd then answer is zero

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

Solution:

\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x                                                 \left[\int_{-a}^{a} f(x) d x=0, f(x)->o d d\right]

\begin{aligned} &f(x)=\cos x \log \left(\frac{1+x}{1-x}\right) \\ & \end{aligned}

f(-x)=\cos (-x) \log \left(\frac{1+(-x)}{1-(-x)}\right)

\begin{aligned} &=\cos x \log \left(\frac{1-x}{1+x}\right) \\ & \end{aligned}

f(-x)=-f(x)

F(x) is odd function

\int_{\frac{-1}{2}}^{\frac{1}{2}} \cos x \log \left(\frac{1+x}{1-x}\right) d x=0

 

 

 

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