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

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

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

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

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 $$$f{\left(x \right)} = \pi \left(\cos{\left(2 x \right)} + 1\right)$$$:

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

Expand the expression:

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

Integrate term by term:

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

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

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

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

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

Let $$$u=2 x$$$.

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

The integral becomes

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

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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

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

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

Recall that $$$u=2 x$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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