Q114 / 160 · TS EAMCET 2023 (14 May Shift 2) · Mathematics

$\int_0^{\frac{\pi}{2}} \frac{x \tan x \sec ^2}{\tan ^4 x+1} d x=$

  • A
    $\frac{\pi^2}{16}$
  • B
    $\frac{\pi^2}{4}$
  • C
    $\frac{\pi^2}{8}$
  • Correct answer
    D
    $\frac{\pi^2}{32}$

Solution

Let $I=\int_0^{\frac{\pi}{2}} \frac{x \tan x \sec ^2 x}{\tan ^4 x+1} d x$ ...(i) $=\int_0^{\frac{\pi}{2}}\left(\frac{\pi}{2}-x\right) \frac{\cot x \cdot \operatorname{cosec}^2 x}{\cot ^4 x+1} d x$ $I=\int_0^\pi \frac{\pi\left(\frac{\pi}{2}-x\right) \tan x \cdot \sec ^2 x}{\tan ^4 x+1} d x$ ...(ii) $2 I=\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \frac{\tan x \cdot \sec ^2 x}{\left(\tan ^2 x\right)^2+1} d x$ Let $\tan ^2 \mathrm{x}=\mathrm{t} \Rightarrow 2 \tan \mathrm{x} \cdot \sec ^2 \mathrm{x} d \mathrm{x}=\mathrm{dt}$ $I=\frac{\pi}{8} \int_0^1 \frac{1}{1+t^2} d t=\frac{\pi}{8}\left[\tan ^{-1} t\right]_0^1=\frac{\pi^2}{32}$

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

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