Q63 / 100 · BITSAT 2024 (Memory Based Paper 1) · 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 answerB$\frac{\pi}{2 \sqrt{2}}$
- C$\frac{\pi}{3 \sqrt{2}}$
- D$\frac{\pi}{4 \sqrt{2}}$
Solution
Let $I=\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x$
$\Rightarrow I=\int_0^{\pi / 2} \frac{\sin \left(\frac{3 \pi}{4}-x\right)+\sin \left(\frac{5 \pi}{4}-x\right)}{\sin x+\cos x} d x$
$\Rightarrow I=\int_0^{\pi / 2} \frac{\sin \left(\frac{\pi}{4}+x\right)-\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x$
Now $I+I=\int_0^{\pi / 2} \frac{2 \sin \left(\frac{\pi}{4}+x\right)}{\cos x+\sin x} d x$
$\Rightarrow I=\int_0^{\pi / 2} \frac{\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x}{\cos x-\sin x} d x=\frac{\pi}{2 \sqrt{2}}$
Hence, the correct answer is option: 'B'.