To integrate \(\int {{{\sin }^2}} x\,dx\) and \(\int {{{\cos }^2}} x\,dx\), a substitution is required.
\(\cos 2x = 1 - 2{\sin ^2}x \to {\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right)\)
\(\cos 2x = 2{\cos ^2}x - 1 \to {\cos ^2}x = \frac{1}{2}\left( {1 + \cos 2x} \right)\)
Hence
\(\begin{align*}\int {{{\sin }^2}} x\,dx &= \frac{1}{2}\int {\left( {1 - \cos 2x} \right)} \,dx\\ &= \frac{1}{2}\left( {x - \frac{1}{2}\sin 2x} \right) + c\end{align*}\)
\(\begin{align*}\int {{{\cos }^2}} x\,dx &= \frac{1}{2}\int {\left( {1 + \cos 2x} \right)} \,dx\\\, &= \frac{1}{2}\left( {x + \frac{1}{2}\sin 2x} \right) + c\end{align*}\)
Integration of the type - \(\int {f'\left( x \right)} {\left( {f(x)} \right)^n}dx\)
For example: Find \(\int {\cos x{{\sin }^4}x\,dx} \)
Let \(u = \sin x \to \frac{{du}}{{dx}} = \cos x \to du = \cos x\,dx\)
\(\begin{align*}\int {\cos x{{\sin }^4}x\,dx} &= \int {{{\sin }^4}x\cos x\,dx} \,\\\, &= \int {{u^4}\,du\,} \\ &= \frac{1}{5}{u^5} = \frac{1}{5}{\sin ^5}x\, + c\end{align*}\)