MHT CET · Maths · Definite Integration
\(\int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} d x=\)
- A \(\frac{\pi}{6}\)
- B \(\frac{\pi}{8}\)
- C \(\frac{\pi}{2}\)
- D \(\frac{\pi}{4}\)
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 \(x=\frac{\pi}{2}, t=1\)
\(\therefore I=\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{1+\mathrm{t}^{2}}=\frac{1}{2}\left[\tan ^{-1} \mathrm{t}\right]_{0}^{1}=\frac{1}{2}\left[\frac{\pi}{4}-0\right]=\frac{\pi}{8}\)
Put \(\sin ^{2} x=t \Rightarrow 2 \sin x \cos x d x=d t\)
When \(x=0, t=0\) and when \(x=\frac{\pi}{2}, t=1\)
\(\therefore I=\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{1+\mathrm{t}^{2}}=\frac{1}{2}\left[\tan ^{-1} \mathrm{t}\right]_{0}^{1}=\frac{1}{2}\left[\frac{\pi}{4}-0\right]=\frac{\pi}{8}\)
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