Question

If ${\int }_0^{\frac{\pi }{2}}{x}^n\sin xdx=\left(\frac{3}{4}\right)({\pi }^2-8)$. Find n

Answer 1

Consider $L.H.S$

$I_n={\int }_0^{\frac{\pi }{2}}{x}^n\sin xdx$

Evaluating by integration by parts we get

$I_n=x^n{\int }_0^{\frac{\pi }{2}}\sin xdx-\int \left({\int }_0^{\frac{\pi }{2}}\sin xdx\right)\left(\frac{d}{dx}{x}^n\right)dx{I}_n={[-x^n\cos x]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}n{x}^{n-1}\cos xdx{I}_n=-[{\left(\frac{\pi }{2}\right)}^n\cos \frac{\pi }{2}-0^n\cos 0]+n{\int }_0^{\frac{\pi }{2}}{x}^{n-1}\cos xdx{I}_n=n{\int }_0^{\frac{\pi }{2}}{x}^{n-1}\cos xdx{I}_n=n[x^{n-1}\int \cos xdx-\int (\int n\cos xdx\frac{d}{dx}(x^{n-1}\left)\right)dx]I_n=n[{\left(x^{n-1}\sin x\right)}_0^{\frac{\pi }{2}}-n(n-1\left){\int }_0^{\frac{\pi }{2}}{x}^{n-2}\sin xdx\right]I_n=n[{\left(\frac{\pi }{2}\right)}^{n-1}\sin \frac{\pi }{2}-0]-(n-1)n{\int }_0^{\frac{\pi }{2}}{x}^{n-2}\sin xdx{I}_n=n{\left[\frac{\pi }{2}\right]}^{n-1}-n(n-1)I_{n-2}$

Now put $n=3$

$I_3=3{\left(\frac{\pi }{2}\right)}^2-6I_1=\frac{3{\pi }^2}{4}-6I_1$

Now to find $I_1$

$I_1={\int }_0^{\frac{\pi }{2}}x\sin xdx=x{\int }_0^{\frac{\pi }{2}}\sin xdx-{\int }_0^{\frac{\pi }{2}}\sin xdx=x{\cos x}_0^{\frac{\pi }{2}}+{\sin x}_0^{\frac{\pi }{2}}=(\frac{\pi }{2}\cos \frac{\pi }{2}+0.\cos 0)+(\sin \frac{\pi }{2}-\sin 0)=1I_3=\frac{3{\pi }^2}{4}-6=\frac{3}{4}({\pi }^2-\frac{24}{3})=\frac{3}{4}({\pi }^2-8)$

So comparing this result with out given equation we can say that $n=3$