MCQ
The correct evaluation of $\int_0^{\pi /2} {\sin x\,\sin 2x} $ is
- A$\frac{4}{3}$
- B$\frac{1}{3}$
- C$\frac{3}{4}$
- ✓$\frac{2}{3}$
$= 2\int_0^{\pi /2} {{{\sin }^2}x\cos xdx} $
Put $t = \sin x \Rightarrow dt = \cos x\,dx$
Now, $I = 2\int_0^1 {{t^2}dt = \frac{2}{3}[{t^3}]_0^1 = \frac{2}{3}} $.
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