Q116 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics

$\int_0^{\pi / 2} \frac{1}{1+\sqrt{\tan x}} d x=$

  • A
    0
  • B
    $\frac{\pi}{2}$
  • C
    $\frac{\pi}{3}$
  • Correct answer
    D
    $\frac{\pi}{4}$

Solution

$\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{\tan x}} d x$ $\qquad ....\mathrm{(i)}$ $I=\int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{\tan \left(\frac{\pi}{2}-x\right)}} d x$ $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{1}{1+\sqrt{\cot x}} d x=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} d x$ $\qquad ....\mathrm{(ii)}$
Adding (i) and (ii) $2 \mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}} d x=\frac{\pi}{2} \Rightarrow \mathrm{I}=\frac{\pi}{4}$

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

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