Q115 / 160 · AP EAMCET 2023 (15 May Shift 2) · Mathematics

$\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x=$

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

Solution

$\begin{aligned} & \text { (b) Let } I \int \frac{-\sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}+x\right)}{\cos x \sin x} d x \\ & \Rightarrow I=\int \frac{12 \sin \left(\frac{3 \pi}{4}-x\right) \sin \left(\frac{5 \pi}{4}-x\right)}{\sin x \cos x} d x \\ & \Rightarrow I=\int_0^{\pi / 2} \frac{\sin \left(\frac{\pi}{4}+x\right)-\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x \\ & \text { Now } I+I=\int_0^{\pi / 2} \frac{2 \sin \left(\frac{\pi}{4}+x\right)}{\cos x+\sin x} d x \\ & \Rightarrow I=\int_0^{\pi / 2} \frac{\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x}{\cos x-\sin x} d x=\frac{\pi}{2 \sqrt{2}} \\ & \end{aligned}$

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

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