Question

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

• A. $\frac{2}{15}$
• B. $\frac{4}{15}$
• C. $\frac{2}{5}$
• D. $0$

Answer 1

Answer: (D)

$0$

Consider, $I={\int }_{-\pi /2}^{\pi /2}\sin x{\cos }^{2}x({\sin }^{2}x+\cos x)dx$

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

consider, $f\left(x\right)={\sin }^{3}x{\cos }^{2}x+\sin x{\cos }^{3}x$

$f(-x)={\sin }^3(-x){\cos }^2(-x)+\sin (-x){\cos }^3(-x)$

$f(-x)=-{\sin }^{3}x{\cos }^{2}x-\sin x{\cos }^{3}x$

$f(-x)=-({\sin }^{3}x{\cos }^{2}x+\sin x{\cos }^{3}x)$

$=-f\left(x\right)$

$\{{\sin }^{3}x{\cos }^{2}x+\sin x{\cos }^{3}x\}$ is an odd function.

Now, we know that for an odd function $F\left(x\right)$,

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

Hence,$I={\int }_{-\pi /2}^{\pi /2}\sin x{\cos }^{2}x({\sin }^{2}x+\cos x)dx=0$