Mathematics - Definite Integration Question with Solution | TestHub

Let Integrating by parts, choosing $\sin x$ as the second function, we get:

$$I_n={\left[x^n(-\cos x)\right]}_0^{\pi /2}-{\int }_0^{\pi /2}n{x}^{n-1}(-\cos x)dx$$

$$=0-n{\int }_0^{\pi /2}{x}^{n-1}\cos xdx$$

Again integrating by parts:

$$I_n=n{\left[x^{n-1}\sin x\right]}_0^{\pi /2}-n(n-1){\int }_0^{\pi /2}{x}^{n-2}\sin xdx$$

$$\Rightarrow I_n=n{\left(\frac{\pi }{2}\right)}^{n-1}-n(n-1)I_{n-2}$$

R.H.S. contains ${\pi }^2$. So, putting $n=3$, we get:

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

$$=\frac{3{\pi }^2}{4}-6{\int }_0^{\pi /2}x\sin xdx$$

$$=\frac{3{\pi }^2}{4}-6[x(-\cos x)+\sin x{]}_0^{\pi /2}$$

$$=\frac{3{\pi }^2}{4}-6(1)=\frac{3}{2}({\pi }^2-8)$$

Hence, $n=3$.