CUET · MATHS · PYQ PAPER 2025
\(\int_0^{\pi / 2} \frac{\sin ^8 x}{\sin ^8 x+\cos ^8 x} d x\) is equal to
- A \(\frac{\pi}{2}\)
- B \(\frac{\pi}{4}\)
- C 0
- D \(-\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
Let \(I = \int_0^{\pi / 2} \frac{\sin ^8 x}{\sin ^8 x+\cos ^8 x} d x\). Using \(\int_0^a f(x) dx = \int_0^a f(a-x) dx\): \(I = \int_0^{\pi / 2} \frac{\sin ^8 (\pi/2 - x)}{\sin ^8 (\pi/2 - x)+\cos ^8 (\pi/2 - x)} d x = \int_0^{\pi / 2} \frac{\cos ^8 x}{\cos ^8 x+\sin ^8 x} d x\).…
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