MCQ
$\int_{ - \pi /2}^{\pi /2} {{{\sin }^2}x{{\cos }^2}x(\sin x + \cos x)\,dx = } $
- A$\frac{2}{{15}}$
- ✓$\frac{4}{{15}}$
- C$\frac{6}{{15}}$
- D$\frac{8}{{15}}$
$= \int_{ - \pi /2}^{\pi /2} {{{\sin }^3}x{{\cos }^2}xdx + \int_{ - \pi /2}^{\pi /2} {{{\sin }^2}x{{\cos }^3}x\,dx} } $
$ = 0 + 2\int_0^{\pi /2} {{{\sin }^2}x{{\cos }^3}xdx} $
$ = 0 + 2 \times \frac{2}{{15}} = \frac{4}{{15}}$ .
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