Q114 / 160 · AP EAMCET 2024 (22 May Shift 1) · Mathematics

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

  • A
    $\pi \log 2+1$
  • B
    $\frac{\pi}{2} \log 2+1$
  • C
    $\frac{\pi}{4} \log 2$
  • Correct answer
    D
    $\frac{\pi}{8} \log 2$

Solution

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

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

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