Answer:

(a) Let \(u=\cos\theta\). Then \(\mathrm{d}u=-\sin\theta\,\mathrm{d}\theta\), so

\(\int \sin\theta\cos^2\theta\,\mathrm{d}\theta = -\int u^2\,\mathrm{d}u = -\frac{u^3}{3}+c = -\frac{1}{3}\cos^3\theta + c.\)

(b) Start with

\(I_n=\int_0^{\pi/2}\sin^n\theta\cos^2\theta\,\mathrm{d}\theta.\)

Use integration by parts in the form \(\sin^n\theta\cos^2\theta=\sin^{n-1}\theta\,(\sin\theta\cos^2\theta)\), with

\(u=\sin^{n-1}\theta,\qquad \mathrm{d}v=\sin\theta\cos^2\theta\,\mathrm{d}\theta.\)

From part (a), \(v=-\frac{1}{3}\cos^3\theta\), and \(\mathrm{d}u=(n-1)\sin^{n-2}\theta\cos\theta\,\mathrm{d}\theta\). Therefore

\(I_n=\left[-\frac{1}{3}\sin^{n-1}\theta\cos^3\theta\right]_0^{\pi/2}+\frac{n-1}{3}\int_0^{\pi/2}\sin^{n-2}\theta\cos^4\theta\,\mathrm{d}\theta.\)

The boundary term is zero, since \(\sin 0=0\) and \(\cos(\pi/2)=0\). Also, \(\cos^4\theta=\cos^2\theta(1-\sin^2\theta)\), so

\(I_n=\frac{n-1}{3}\int_0^{\pi/2}\sin^{n-2}\theta\cos^2\theta(1-\sin^2\theta)\,\mathrm{d}\theta.\)

Hence

\(I_n=\frac{n-1}{3}\left(\int_0^{\pi/2}\sin^{n-2}\theta\cos^2\theta\,\mathrm{d}\theta-\int_0^{\pi/2}\sin^n\theta\cos^2\theta\,\mathrm{d}\theta\right).\)

So

\(I_n=\frac{n-1}{3}(I_{n-2}-I_n).\)

Rearranging,

\(3I_n=(n-1)I_{n-2}-(n-1)I_n\)

\(\Rightarrow (n+2)I_n=(n-1)I_{n-2}\)

\(\Rightarrow I_n=\frac{n-1}{n+2}I_{n-2}.\)

Now take \(n=4\):

\(I_4=\frac{3}{6}I_2=\frac12 I_2.\)

Next, \(I_2=\int_0^{\pi/2}\sin^2\theta\cos^2\theta\,\mathrm{d}\theta\). It is simpler to start from \(I_0\):

\(I_0=\int_0^{\pi/2}\cos^2\theta\,\mathrm{d}\theta=\frac12\int_0^{\pi/2}(1+\cos 2\theta)\,\mathrm{d}\theta=\frac12\left[\theta+\frac12\sin 2\theta\right]_0^{\pi/2}=\frac{\pi}{4}.\)

Then

\(I_2=\frac{1}{4}I_0=\frac{\pi}{16}\)

and therefore

\(I_4=\frac12\cdot\frac{\pi}{16}=\frac{\pi}{32}.\)