Q87 / 150 · MHT CET 2011 · Mathematics

$\int_{0}^{\pi / 2} \frac{d x}{1+\tan x}$ is equal to

  • A
    $\pi$
  • B
    $\pi / 2$
  • C
    $\pi / 3$
  • Correct answer
    D
    $\pi / 4$

Solution

Given, $I=\int_{0}^{\pi / 2} \frac{d x}{1+\tan x}$ $$ \begin{array}{rlr} I & =\int_{0}^{\pi / 2} \frac{\cos x}{\sin x+\cos x} d x & \ldots \\ I & =\int_{0}^{\pi / 2} \frac{\cos (\pi / 2-x)}{\sin (\pi / 2-x)+\cos (\pi / 2-x)} d x \\ & =\int_{0}^{\pi / 2} \frac{\sin x}{\cos x+\sin x} d x & \ldots \end{array} $$ On adding Eqs. (i) and (ii), we get $$ \begin{aligned} 2 I &=\int_{0}^{\pi / 2}\left(\frac{\sin x+\cos x}{\sin x+\cos x}\right) d x \\ &=\int_{0}^{\pi / 2} d x=\pi / 2 \\ \Rightarrow \quad I &=\pi / 4 \end{aligned} $$

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

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