Q114 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics

$\int_{-1 / 24}^{1 / 24} \sec x \log \left(\frac{1-x}{1+x}\right) d x=$

  • A
    $\frac{\pi}{2}$
  • B
    $\pi$
  • C
    1
  • Correct answer
    D
    0

Solution

$I=\int_{\frac{-1}{24}}^{\frac{1}{24}} \sec \log \left(\frac{1-x}{1+x}\right) d x$ $\begin{aligned} & f(x)=\sec x \log \left(\frac{1-x}{1+x}\right) \\ & f(-x)=\sec (-x) \log \left(\frac{1+x}{1-x}\right) \\ & f(-x)=\sec (x) \log \left(\frac{1-x}{1+x}\right)^{-1} \\ & f(-x)=-\sec (x) \log \left(\frac{1-x}{1+x}\right) \\ & \Rightarrow f(-x)=-f(x) \\ & \therefore \mathrm{I}=0\end{aligned}$

Hence, the correct answer is option: 'D'.

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