COMEDK · Maths · 30. Definite Integration
\(\int_{0}^{\pi / 2} \frac{\sin 2 t}{\sin ^{4} t+\cos ^{4} t} d t=\)
- A \(\pi\)
- B \(\frac{\pi}{3}\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} I &=\int_{0}^{\pi / 2} \frac{\sin 2 t}{\sin ^{4} t+\cos ^{4} t} d t \\ &=\int_{0}^{\pi / 2} \frac{2 \sin t \cos t}{\sin ^{4} t+\cos ^{4} t} d t \\ &=\int_{0}^{\pi / 2} \frac{2 \tan t \sec ^{2} t}{\left(\tan ^{2} t\right)^{2}+1} d t \end{aligned} \]…
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