Q211 / 270 · NDA 2022 (Phase 1) · Mathematics

What is $\int_0^{\frac{\pi}{4}} \frac{d x}{(\sin x+\cos x)^2}$ equal to?

  • A
    $-\frac{1}{2}$
  • Correct answer
    B
    $\frac{1}{2}$
  • C
    1
  • D
    $\frac{3}{2}$

Solution

$\begin{aligned} & \int_0^{\frac{\pi}{4}} \frac{d x}{(\sin x+\cos x)^2}=\frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{4}} \frac{d x}{\left(\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos \right)^2} \\ = & \frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{4}} \frac{d x}{\left(\operatorname{cas}^2\left(\frac{\pi}{4}-x\right)\right.}=\frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{4}} \sec ^2\left(\frac{\pi}{4} x\right) d x \\ = & \frac{1}{\sqrt{2}}\left[-\tan \left(\frac{\pi}{4}-x\right)\right]_0^{\frac{\pi}{4}}=\frac{-1}{\sqrt{2}}\left[\tan 0-\tan \frac{\pi}{4}\right] \\ = & \frac{-1}{\sqrt{2}}\left[0-\frac{1}{\sqrt{2}}\right]=\frac{1}{2}\end{aligned}$

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

Relevant Questions