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
    $\frac{8}{{15}}$

Answer

Correct option: B.
$\frac{4}{{15}}$
b
(b) $\int_{ - \pi /2}^{\pi /2} {{{\sin }^2}x{{\cos }^2}x(\sin x + \cos x)dx} $

$= \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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