Integration using Double Angle Formula
$$ \large \begin{align} \displaystyle
\cos{2x} &= 2\cos^2{x}-1 \\
&= 1-2 \sin^2 {x} \\
&= \cos^2{x}-\sin^2{x} \\
\sin{2x} &= 2 \sin{x} \cos{x}
\end{align} $$
These will be expressed in the following forms to apply to integration.
$$ \large \begin{align} \displaystyle
\sin^2 x &= \dfrac{1}{2}(1-\cos 2x) \\
\cos^2 x &= \dfrac{1}{2}(1 + \cos 2x) \\
\sin x \cos x &= \dfrac{1}{2} \sin 2x
\end{align} $$
Example 1
Find $\displaystyle \int{\sin^2{x}}dx$.
\( \begin{align} \displaystyle
\cos{2x} &= 1-2 \sin^2{x} \\
2 \sin^2{x} &= 1-\cos{2x} \\
\sin^2{x} &= \dfrac{1}{2}(1-\cos{2x}) \\
\int{\sin^2{x}}dx &= \dfrac{1}{2} \int{(1-\cos{2x})}dx \\
&= \dfrac{1}{2}\big(x-\dfrac{1}{2}\sin{2x}\big)+c
\end{align} \)
Example 2
Find $\displaystyle \int{\cos^2{x}}dx$.
\( \begin{align} \displaystyle
\cos{2x} &= 2 \cos^2{x}-1 \\
2 \cos^2{x} &= 1 + \cos{2x} \\
\cos^2{x} &= \dfrac{1}{2}(1 + \cos{2x}) \\
\int{\cos^2{x}}dx &= \dfrac{1}{2} \int{(1+ \cos{2x})}dx \\
&= \dfrac{1}{2}\big(x+\dfrac{1}{2}\sin{2x}\big) + c
\end{align} \)
Example 3
Find $\displaystyle \int{\sin^2{x}}dx$.
\( \begin{align} \displaystyle
\cos{2x} &= 1-2 \sin^2{x} \\
2 \sin^2{x} &= 1-\cos{2x} \\
\sin^2{x} &= \dfrac{1}{2}(1-\cos{2x}) \\
\sin^2{2x} &= \dfrac{1}{2}(1-\cos{4x}) \\
\int{\sin^2{2x}}dx &= \dfrac{1}{2} \int{(1-\cos{4x})}dx \\
&= \dfrac{1}{2}\big(x-\dfrac{1}{4}\sin{4x}\big)+c
\end{align} \)
Example 4
Find $\displaystyle \int{\cos^2{6x}}dx$.
\( \begin{align} \displaystyle
\cos{2x} &= 2 \cos^2{x}-1 \\
2 \cos^2{x} &= 1 + \cos{2x} \\
\cos^2{x} &= \dfrac{1}{2}(1 + \cos{2x}) \\
\cos^2{6x} &= \dfrac{1}{2}(1 + \cos{12x}) \\
\int{\cos^2{6x}}dx &= \dfrac{1}{2} \int{(1+ \cos{12x})}dx \\
&= \dfrac{1}{2}\big(x+\dfrac{1}{12}\sin{12x}\big) + c
\end{align} \)
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