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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

Answers (1)

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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