MCQ
$\int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^4 x} d x=$
- A$\frac{\pi}{2}$
- B$\frac{\pi}{4}$
- C$\frac{\pi}{6}$
- ✓$\frac{\pi}{8}$
✓
Answer
Correct option: D.
$\frac{\pi}{8}$
(D)
Put $\sin ^2 x= t \Rightarrow 2 \sin x \cos x d x= dt$
When $x=0, t =0$ and when $x=\frac{\pi}{2}, t =1$
$\therefore \int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^4 x} d x=\frac{1}{2} \int_0^1 \frac{1}{1+ t ^2} dt =\frac{1}{2}\left[\tan ^{-1} t \right]_0^1=\frac{\pi}{8}$
Put $\sin ^2 x= t \Rightarrow 2 \sin x \cos x d x= dt$
When $x=0, t =0$ and when $x=\frac{\pi}{2}, t =1$
$\therefore \int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^4 x} d x=\frac{1}{2} \int_0^1 \frac{1}{1+ t ^2} dt =\frac{1}{2}\left[\tan ^{-1} t \right]_0^1=\frac{\pi}{8}$
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