Question

Evaluate the following integrals:

${\int }_0^{\mathrm{\pi }}x\sin x{\cos }^{2}xdx$ [NCERT EXEMPLAR]

Answer 1

Let I = ${\int }_0^{\mathrm{\pi }}x\sin x{\cos }^{2}xdx$ .....(1)
Then,
$$\begin{gathered} I={\int }_0^{\mathrm{\pi }}\left(\mathrm{\pi }-x\right)\sin \left(\mathrm{\pi }-x\right){\cos }^2\left(\mathrm{\pi }-x\right)dx\left[{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right] \\ ={\int }_0^{\mathrm{\pi }}\left(\mathrm{\pi }-x\right)\sin x{\cos }^{2}xdx.....\left(2\right) \end{gathered}$$

Adding (1) and (2), we have

$$\begin{gathered} 2I={\int }_0^{\mathrm{\pi }}\left(\mathrm{\pi }-x+x\right)\sin x{\cos }^{2}xdx \\ \Rightarrow 2I=\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\sin x{\cos }^{2}xdx \\ \Rightarrow 2I=-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}{\cos }^{2}x\left(-\sin x\right)dx \\ \Rightarrow 2I=-\mathrm{\pi }\times {\overline{)\frac{{\cos }^{3}x}{3}}}_0^{\mathrm{\pi }}\left[\int {\left[f\left(x\right)\right]}^{n}f\text{'}\left(x\right)dx=\frac{{\left[f\left(x\right)\right]}^{n+1}}{n+1}+C\right] \\ \Rightarrow 2I=-\frac{\mathrm{\pi }}{3}\left({\cos }^3\mathrm{\pi }-{\cos }^{2}0\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow 2I=-\frac{\mathrm{\pi }}{3}\left(-1-1\right)=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow I=\frac{\mathrm{\pi }}{3} \end{gathered}$$