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MHT CET · Maths · Definite Integration

\(\int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} d x=\)

  1. A \(\frac{\pi}{6}\)
  2. B \(\frac{\pi}{8}\)
  3. C \(\frac{\pi}{2}\)
  4. D \(\frac{\pi}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{\pi}{8}\)

Step-by-step Solution

Detailed explanation

\(I=\int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} d x \quad=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{2 \sin x \cos x}{1+\left(\sin ^{2} x\right)^{2}} d x\)
Put \(\sin ^{2} x=t \Rightarrow 2 \sin x \cos x d x=d t\)
When \(x=0, t=0\) and when \(x=\frac{\pi}{2}, t=1\)
\(\therefore I=\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{1+\mathrm{t}^{2}}=\frac{1}{2}\left[\tan ^{-1} \mathrm{t}\right]_{0}^{1}=\frac{1}{2}\left[\frac{\pi}{4}-0\right]=\frac{\pi}{8}\)