integral exercise asked by Kalipada Das in math groups (Reduction Formula _ BY Parts)

Exercise:

Let ${{I}_{n}}=\int_{0}^{\frac{\pi }{2}}{{{x}^{n}}\sin x}\,dx$

Show that, ${{I}_{n}}+n\left( n-1 \right){{I}_{n-2}}=n{{\left( \frac{\pi }{2} \right)}^{n-1}}$ and deduce the value of $\int_{0}^{\frac{\pi }{2}}{{{x}^{5}}\sin x}\,dx$

Solution: Let $u={{x}^{n}}\,\,\And \,\,\,dv=\sin x\,dx\Leftrightarrow du=n{{x}^{n-1}}dx\,\,\And v=\cos x$

${{I}_{n}}=\int_{0}^{\frac{\pi }{2}}{{{x}^{n}}\sin x\,dx}=\left( {{x}^{n}}\cos x \right)_{0}^{\frac{\pi }{2}}-n\int_{0}^{\frac{\pi }{2}}{{{x}^{n-1}}\cos xdx}$

Let $u={{x}^{n-1}}\,\,\And \,\,\,dv=\cos x\,dx\Leftrightarrow du=\left( n-1 \right){{x}^{n-2}}dx\,\,\And \,\,v=-\sin x\,$

So $\int_{0}^{\frac{\pi }{2}}{{{x}^{n-1}}\cos x\,dx}=-\left( {{x}^{n-1}}\sin x \right)_{0}^{\frac{\pi }{2}}+\left( n-1 \right)\int_{0}^{\frac{\pi }{2}}{{{x}^{n-2}}\sin x\,dx}$

So ${{I}_{n}}=\left( {{x}^{n}}\cos x \right)_{0}^{\frac{\pi }{2}}+n\left( {{x}^{n-1}}\sin x \right)_{0}^{\frac{\pi }{2}}-n\left( n-1 \right){{I}_{n-2}}$

$\Leftrightarrow {{I}_{n}}+n\left( n-1 \right){{I}_{n-2}}=\left( {{x}^{n}}\cos x \right)_{0}^{\frac{\pi }{2}}+n\left( {{x}^{n-1}}\sin x \right)_{0}^{\frac{\pi }{2}}=n{{\left( \frac{\pi }{2} \right)}^{n-1}}$

Thus $\int_{0}^{\frac{\pi }{2}}{{{x}^{5}}\sin x\,dx}=5{{\left( \frac{\pi }{2} \right)}^{5-1}}=\frac{5{{\pi }^{4}}}{16}$