(i) $\int_0^{\pi / 2} \sin ^m x \cdot \cos x d x$
Put, $\sin x=t$
$$
\begin{aligned}
\cos x d x & =d t=\int_0^{\pi / 2} t^m d t \\
& =\left[\frac{t^{m+1}}{m+1}\right]_0^{\pi / 2}=\left[\frac{(\sin x)^{m+1}}{m+1}\right]_0^{\pi / 2}
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{1}{m+1}\left[(\sin x)^{m+1}\right]_0^{\pi / 2} \\
& =\frac{1}{m+1}\left[\sin ^{m+1}\left(\frac{\pi}{2}\right)-\sin ^{m+1} \cdot(0)\right] \\
& =\frac{1}{m+1}[(1)-(0)]=\frac{1}{m+1}
\end{aligned}
$$
(ii) $\int_0^{\pi / 2} \sin x \cdot \cos ^n(x)=d x$
$$
\begin{aligned}
\cos x & =t \\
-\sin x d x & =d t \\
\sin x \cdot d x & =-d t \\
\int_0^{\pi / 2} t^n \cdot(-d t) & =-\int_0^{\pi / 2} t^n \cdot d t \\
= & -\left[\frac{t^{n+1}}{n+1}\right]_0^{\pi / 2}=-\left[\frac{\cos x^{n+1}}{n+1}\right]_0^{\pi / 2} \\
& =\frac{-1}{n+1}\left[\cos ^{n+1} x\right]_0^{\pi / 2}
\end{aligned}
$$


$$
\begin{aligned}
& =\frac{-1}{n+1}\left[\cos ^{n+1} \frac{\pi}{2}-\cos ^{n+1}(0)\right] \\
& =\frac{-1}{n+1}[0-1]=\frac{1}{n+1}
\end{aligned}
$$
Hence, option (4) is correct.