Question

$\underset{-\pi /2}{\overset{\pi /2}{\int }}{\sin }^{2}x{\cos }^{2}x(\sin x+\cos x)dx$ is equal to

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

Answer 1

Answer: (C)

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