# 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}