JEE Mains · Maths · STD 12 - 7.2 definite integral
The value of \(\int\limits_{0}^{2 \pi} \frac{x \sin ^{8} x}{\sin ^{8} x+\cos ^{8} x} d x\) is equal to
- A \(2 \pi\)
- B \(4 \pi\)
- C \(2 \pi ^2\)
- D \(4 \pi ^2\)
Answer & Solution
Correct Answer
(D) \(4 \pi ^2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I}=\int_{0}^{2 \pi} \frac{\mathrm{x} \sin ^{8} \mathrm{x}}{\sin ^{8} \mathrm{x}+\cos ^{8} \mathrm{x}} \mathrm{dx}\)…
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