When the integrand involves even powers of sines or cosines, we recall the Double Angle Formulae for cosine:
\[
\begin{align*}
\cos^2 \theta & = \frac{1}{2} (1 + \cos 2\theta) \\
 & = \frac{1}{2} (1 - \cos 2\theta).
\end{align*}
\]
Proof.
Recall that:
\[
\cos 2\theta = 2 \cos^2 \theta - 1
\]
Rearranging, we have \[
\begin{align*}
2 \cos^2 \theta & = 1 + \cos 2\theta \\
\cos^2 \theta & = \frac{1}{2} (1 + 2 \cos 2\theta).
\end{align*}
\]
Similarly, for sine, \[
\cos 2\theta = 1 - 2 \sin^2 \theta.
\]
So, we have \[
\begin{align*}
2 \sin^2 \theta & = 1 - \cos 2\theta \\
\sin^2 \theta & = \frac{1}{2} (1- \cos 2\theta).
\end{align*}
\]
\(\Box\)

When the integrand involves even powers of sines or cosines, we recall the Double Angle Formulae for cosine:

\[
\begin{align*}
\cos^2 \theta & = \frac{1}{2} (1 + \cos 2\theta) \\
                             & = \frac{1}{2} (1 - \cos 2\theta).
\end{align*}
\]

Rearranging, we have
\[
\begin{align*}
2 \cos^2 \theta & = 1 + \cos 2\theta \\
\cos^2 \theta & = \frac{1}{2} (1 + 2 \cos 2\theta).
\end{align*}
\]

Similarly, for sine,
\[
\cos 2\theta = 1 - 2 \sin^2 \theta.
\]

So, we have
\[
\begin{align*}
2 \sin^2 \theta & = 1 - \cos 2\theta \\
\sin^2 \theta & = \frac{1}{2} (1- \cos 2\theta).
\end{align*}
\]

Integrable Forms of Trigonometric Functions

Techniques of  Integration

International Baccalaureate Higher Level Mathematics

Techniques of Integration - Integrable Forms of Trigonometric Functions

Techniques of Integration - Integrable Forms of Trigonometric Functions