Answer: (i) For \(I_n=\int_0^{\frac{\pi}{2}} x^n\sin x\,dx\), integrate by parts with \(u=x^n\) and \(dv=\sin x\,dx\). Then \(du=nx^{n-1}dx\) and \(v=-\cos x\).
So \(I_n=\left[-x^n\cos x\right]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}} nx^{n-1}\cos x\,dx\).
Now integrate the remaining integral by parts again, with \(u=nx^{n-1}\) and \(dv=\cos x\,dx\). Then \(du=n(n-1)x^{n-2}dx\) and \(v=\sin x\), giving
\(I_n=\left[-x^n\cos x\right]_0^{\frac{\pi}{2}}+\left[nx^{n-1}\sin x\right]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} n(n-1)x^{n-2}\sin x\,dx.\)
Since \(\cos\frac{\pi}{2}=0\), \(\cos 0=1\), and \(x^n=0\) at \(x=0\), the first bracket is \(0\). Also \(\sin\frac{\pi}{2}=1\) and \(\sin 0=0\), so the second bracket is \(n\left(\frac{\pi}{2}\right)^{n-1}\). The integral is \(n(n-1)I_{n-2}\). Hence
\(I_n=n\left(\frac{\pi}{2}\right)^{n-1}-n(n-1)I_{n-2}\),
so rearranging gives \(I_n+n(n-1)I_{n-2}=n\left(\frac{\pi}{2}\right)^{n-1}\).
(ii) For \(I_1\), use integration by parts with \(u=x\), \(dv=\sin x\,dx\):
\(I_1=\int_0^{\frac{\pi}{2}} x\sin x\,dx=\left[-x\cos x\right]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}} \cos x\,dx=\left[\sin x\right]_0^{\frac{\pi}{2}}=1.\)
Now take \(n=3\) in the recurrence:
\(I_3+3\cdot2\,I_1=3\left(\frac{\pi}{2}\right)^2.\)
Since \(I_1=1\),
\(I_3=3\left(\frac{\pi}{2}\right)^2-6=\frac{3\pi^2}{4}-6.\)
(i) Let \(I_n=\int_0^{\frac{\pi}{2}} x^n\sin x\,dx\). Use integration by parts with \(u=x^n\) and \(dv=\sin x\,dx\). Then \(du=nx^{n-1}dx\) and \(v=-\cos x\), so
\(I_n=\left[-x^n\cos x\right]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}} nx^{n-1}\cos x\,dx.\)
Now integrate \(\int_0^{\frac{\pi}{2}} nx^{n-1}\cos x\,dx\) by parts again, taking \(u=nx^{n-1}\) and \(dv=\cos x\,dx\). Then \(du=n(n-1)x^{n-2}dx\) and \(v=\sin x\), giving
\(\int_0^{\frac{\pi}{2}} nx^{n-1}\cos x\,dx=\left[nx^{n-1}\sin x\right]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} n(n-1)x^{n-2}\sin x\,dx.\)
Substitute this into the expression for \(I_n\):
\(I_n=\left[-x^n\cos x\right]_0^{\frac{\pi}{2}}+\left[nx^{n-1}\sin x\right]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} n(n-1)x^{n-2}\sin x\,dx.\)
At \(x=\frac{\pi}{2}\), \(\cos x=0\); at \(x=0\), \(x^n=0\); so the first bracket is \(0\). Also \(\sin\frac{\pi}{2}=1\) and \(\sin 0=0\), so the second bracket is \(n\left(\frac{\pi}{2}\right)^{n-1}\). The remaining integral is exactly \(n(n-1)I_{n-2}\). Hence
\(I_n=n\left(\frac{\pi}{2}\right)^{n-1}-n(n-1)I_{n-2}.\)
Rearranging gives
\(I_n+n(n-1)I_{n-2}=n\left(\frac{\pi}{2}\right)^{n-1}.\)
(ii) For \(I_1\), integrate by parts with \(u=x\), \(dv=\sin x\,dx\):
\(I_1=\int_0^{\frac{\pi}{2}} x\sin x\,dx=\left[-x\cos x\right]_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}} \cos x\,dx.\)
The boundary term is zero, so
\(I_1=\left[\sin x\right]_0^{\frac{\pi}{2}}=1.\)
Now use the recurrence with \(n=3\):
\(I_3+3\cdot2\,I_1=3\left(\frac{\pi}{2}\right)^2.\)
Since \(I_1=1\),
\(I_3=3\left(\frac{\pi}{2}\right)^2-6=\frac{3\pi^2}{4}-6.\)
