Question

$\int\limits_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x$ is equal to

• A. $\frac{2}{15}$
• B. $\frac{1}{5}$
• C. $\frac{4}{15}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

$\int\limits_{-\pi / 2}^{\pi / 2} \sin ^2 x \cos ^2 x(\sin x+\cos x) d x$
$=\int\limits_{-\pi / 2}^{\pi / 2} \underset{\text{(odd function)}}{ \sin ^3 x \cos ^2 x d x}+\int\limits_{-\pi / 2}^{\pi / 2} \underset{\text{(even function)}}{\sin ^2 x \cos ^3 x d x}$
$=0+2 \int\limits_0^{\pi / 2} \sin ^2 x \cos ^3 x d x$
$=2 \cdot \frac{1.2}{5 \cdot 3 \cdot 1}=\frac{4}{15}$