Q58 / 150 · MHT CET 2021 (20 Sep Shift 1) · Mathematics

$\int_0^{\pi / 4} \log (1+\tan x) d x=$

  • A
    $\frac{\pi}{16} \log 2$
  • B
    $\frac{\pi}{4} \log 2$
  • Correct answer
    C
    $\frac{\pi}{8} \log 2$
  • D
    $\pi \log 2$

Solution

$\begin{aligned} & \text { Let } I=\int_0^{\pi / 4} \log (1+\tan x) d x \\ & =\int_0^{\pi / 4} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x \\ & =\int_0^{\pi / 4} \log \left[1+\left(\frac{1-\tan x}{1+\tan x}\right)\right] d x=\int_0^{\pi / 4} \log \left(\frac{2}{1+\tan x}\right) d x \\ & =\int_0^{\pi / 4}(\log 2) d x-\int_0^{\pi / 4} \log (1+\tan x) d x=\int_0^{\pi / 4}(\log 2) d x-I \\ & \therefore 2 I=(\log 2)[x]_0^{\pi / 4}=(\log 2)\left(\frac{\pi}{4}\right) \Rightarrow I=\left(\frac{\pi}{8}\right) \log 2\end{aligned}$

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

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