Question

Evaluate: ${\int }_{-\pi /2}^{\pi /2}{\sin }^{2}x.{\cos }^{2}x(\sin x+\cos x)dx$

• A. $\frac{1}{15}$
• B. $\frac{2}{15}$
• C. $\frac{4}{15}$
• D. $0$

Answer 1

Answer: (C)

$\frac{4}{15}$

${\int }_{-\pi /2}^{\pi /2}{\sin }^{2}x.{\cos }^{2}x(\sin x+\cos x)dx$

$={\int }_{-\pi /2}^{\pi /2}{\sin }^{3}x{\cos }^{2}xdx+{\int }_{-\pi /2}^{\pi /2}{\sin }^{2}x{\cos }^{3}xdx$

Now we know that for an odd function $f\left(x\right)$:-

${\int }_{-a}^{a}f\left(x\right)dx=0$

And for an even function $g\left(x\right)$:-

${\int }_{-a}^{a}g\left(x\right)dx=2{\int }_0^{a}g\left(x\right)dx$

Now, $f\left(x\right)={\sin }^{3}x.{\cos }^{2}x$ is an odd function and $g\left(x\right)={\sin }^{2}x.{\cos }^{3}x$ is an even function.

Hence, first integration is $0$.

For second part,we get-

$2{\int }_0^{\pi /2}{\sin }^{2}x{\cos }^{3}xdx$

$=2{\int }_0^{\pi /2}{\sin }^{2}x(1-{\sin }^{2}x)\cos xdx$

Now, let $t=\sin x$ $⟹dt=\cos xdx$

And for $x=0,t=0$ and for $x=\frac{\pi }{2},t=1$

Hence, integration becomes:-

$=2{\int }_0^1{t}^2(1-t^2)dt=2{\int }_0^1(t^2-t^4)dt$

$=2{[\frac{t^3}{3}-\frac{t^5}{5}]}_0^1$

$=2(\frac{1}{3}-\frac{1}{5})$

$=\frac{4}{15}$

Hence ,answer is option-(C).