Integral of $\sin^{2}{\left(x \right)}$

Find $\int \sin^{2}{\left(x \right)}\, dx$.

Apply the power reducing formula $\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$ with $\alpha=x$:

$${\color{red}{\int{\sin^{2}{\left(x \right)} d x}}} = {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)d x}}}$$

Apply the constant multiple rule $\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$ with $c=\frac{1}{2}$ and $f{\left(x \right)} = 1 - \cos{\left(2 x \right)}$:

$${\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 x \right)}\right)d x}}{2}\right)}}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(1 - \cos{\left(2 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{1 d x} - \int{\cos{\left(2 x \right)} d x}\right)}}}{2}$$

Apply the constant rule $\int c\, dx = c x$ with $c=1$:

$$- \frac{\int{\cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{\int{1 d x}}}}{2} = - \frac{\int{\cos{\left(2 x \right)} d x}}{2} + \frac{{\color{red}{x}}}{2}$$

Let $u=2 x$.

Then $du=\left(2 x\right)^{\prime }dx = 2 dx$ (steps can be seen »), and we have that $dx = \frac{du}{2}$.

Therefore,

$$\frac{x}{2} - \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{2} = \frac{x}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2}$$

Apply the constant multiple rule $\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$ with $c=\frac{1}{2}$ and $f{\left(u \right)} = \cos{\left(u \right)}$:

$$\frac{x}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2} = \frac{x}{2} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{2}$$

The integral of the cosine is $\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$:

$$\frac{x}{2} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = \frac{x}{2} - \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$

Recall that $u=2 x$:

$$\frac{x}{2} - \frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{x}{2} - \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$

Therefore,

$$\int{\sin^{2}{\left(x \right)} d x} = \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}$$

Add the constant of integration:

$$\int{\sin^{2}{\left(x \right)} d x} = \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}+C$$

$\int \sin^{2}{\left(x \right)}\, dx = \left(\frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}\right) + C$A