Question

The value of $\underset{-\mathrm{\pi }}{\overset{\mathrm{\pi }}{\int }}{\sin }^{3}x{\cos }^{2}xdx$ is

(a) $\frac{{\mathrm{\pi }}^4}{2}$

(b) $\frac{{\mathrm{\pi }}^4}{4}$

(c) 0

(d) none of these

Answer 1

(c) 0

$$\begin{gathered} {\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}{\sin }^{3}x{\cos }^{2}xdx \\ ={\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\sin x\left(1-{\cos }^{2}x\right){\cos }^{2}xdx \\ \mathrm{Let}\cos x=t,\mathrm{then}-\sin xdx=dt, \\ \mathrm{When},x=-\mathrm{\pi },\mathrm{t}=-1,\mathrm{x}=\mathrm{\pi },\mathrm{t}=-1 \\ \mathrm{Therefore}\mathrm{the}\mathrm{integral}\mathrm{becomes} \\ {\int }_{-1}^{-1}-\left(1-{\mathrm{t}}^2\right){\mathrm{t}}^2\mathrm{dt} \\ =0 \end{gathered}$$