Integral of $$$\sin^{2}{\left(\frac{\pi m x}{a} \right)}$$$ with respect to $$$x$$$

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

Let $$$u=\frac{\pi m x}{a}$$$.

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

The integral becomes

$${\color{red}{\int{\sin^{2}{\left(\frac{\pi m x}{a} \right)} d x}}} = {\color{red}{\int{\frac{a \sin^{2}{\left(u \right)}}{\pi m} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{a}{\pi m}$$$ and $$$f{\left(u \right)} = \sin^{2}{\left(u \right)}$$$:

$${\color{red}{\int{\frac{a \sin^{2}{\left(u \right)}}{\pi m} d u}}} = {\color{red}{\frac{a \int{\sin^{2}{\left(u \right)} d u}}{\pi m}}}$$

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

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

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 $$$f{\left(u \right)} = 1 - \cos{\left(2 u \right)}$$$:

$$\frac{a {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 u \right)}}{2}\right)d u}}}}{\pi m} = \frac{a {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 u \right)}\right)d u}}{2}\right)}}}{\pi m}$$

Integrate term by term:

$$\frac{a {\color{red}{\int{\left(1 - \cos{\left(2 u \right)}\right)d u}}}}{2 \pi m} = \frac{a {\color{red}{\left(\int{1 d u} - \int{\cos{\left(2 u \right)} d u}\right)}}}{2 \pi m}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$$\frac{a \left(- \int{\cos{\left(2 u \right)} d u} + {\color{red}{\int{1 d u}}}\right)}{2 \pi m} = \frac{a \left(- \int{\cos{\left(2 u \right)} d u} + {\color{red}{u}}\right)}{2 \pi m}$$

Let $$$v=2 u$$$.

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

The integral can be rewritten as

$$\frac{a \left(u - {\color{red}{\int{\cos{\left(2 u \right)} d u}}}\right)}{2 \pi m} = \frac{a \left(u - {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}\right)}{2 \pi m}$$

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

$$\frac{a \left(u - {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}\right)}{2 \pi m} = \frac{a \left(u - {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}\right)}{2 \pi m}$$

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

$$\frac{a \left(u - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{2}\right)}{2 \pi m} = \frac{a \left(u - \frac{{\color{red}{\sin{\left(v \right)}}}}{2}\right)}{2 \pi m}$$

Recall that $$$v=2 u$$$:

$$\frac{a \left(u - \frac{\sin{\left({\color{red}{v}} \right)}}{2}\right)}{2 \pi m} = \frac{a \left(u - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{2}\right)}{2 \pi m}$$

Recall that $$$u=\frac{\pi m x}{a}$$$:

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

Therefore,

$$\int{\sin^{2}{\left(\frac{\pi m x}{a} \right)} d x} = \frac{a \left(- \frac{\sin{\left(\frac{2 \pi m x}{a} \right)}}{2} + \frac{\pi m x}{a}\right)}{2 \pi m}$$

Simplify:

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

Add the constant of integration:

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

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