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. $\frac{4}{15}$
• C. $\frac{6}{15}$
• D. $\frac{8}{15}$

Answer 1

Answer: (B)

$\underset{-\pi /2}{\overset{\pi /2}{\int }}si{n}^{2}xco{s}^{2}x\left(sinx+cosx\right)dx$
$=\underset{-\pi /2}{\overset{\pi /2}{\int }}si{n}^{3}xco{s}^{2}xdx+\underset{-\pi /2}{\overset{\pi /2}{\int }}si{n}^{2}xco{s}^{3}xdx$
[$\because$ $si{n}^{3}xco{s}^{2}x$ is odd and $si{n}^{2}xco{s}^{3}x$ is even]
$=0+2\underset{0}{\overset{\pi /2}{\int }}si{n}^{2}xco{s}^{3}xdx$
Put $sinx=z\therefore cosxdx-dz$
$\therefore$ given integral $=2\underset{0}{\overset{1}{\int }}{z}^2\left(1-z^2\right)dz$
$=2{\left|\frac{z^3}{3}-\frac{z^5}{5}\right|}_0^1$
$=2\left(\frac{1}{3}-\frac{1}{5}\right)=\frac{4}{15}$