MCQ
$\int_0^{\pi /2} {\frac{{\sin x\cos x}}{{1 + {{\sin }^4}x}}\,dx = } $
- A$\frac{\pi }{2}$
- B$\frac{\pi }{4}$
- C$\frac{\pi }{6}$
- ✓$\frac{\pi }{8}$
✓
Answer
Correct option: D.
$\frac{\pi }{8}$
d
(d) Put ${\sin ^2}x = t \Rightarrow dt = 2\sin x\cos x\,dx$
(d) Put ${\sin ^2}x = t \Rightarrow dt = 2\sin x\cos x\,dx$
Now $\int_0^{\pi /2} {\frac{{\sin x\cos x}}{{1 + {{\sin }^4}x}}dx = \frac{1}{2}\int_0^1 {\frac{1}{{1 + {t^2}}}dt = \frac{1}{2}[{{\tan }^{ - 1}}t]_0^1 = \frac{\pi }{8}} } $.
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