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

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

Integrate term by term:

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

Rewrite the sine using the double angle formula $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:

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

Multiply the numerator and denominator by $$$\sec^2\left(\frac{x}{2} \right)$$$:

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

Let $$$u=\tan{\left(\frac{x}{2} \right)}$$$.

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

Therefore,

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

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

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

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$- a^{2} {\color{red}{\int{\frac{1}{u} d u}}} + \int{\sin{\left(x \right)} d x} = - a^{2} {\color{red}{\ln{\left(\left|{u}\right| \right)}}} + \int{\sin{\left(x \right)} d x}$$

Recall that $$$u=\tan{\left(\frac{x}{2} \right)}$$$:

$$- a^{2} \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{\sin{\left(x \right)} d x} = - a^{2} \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} \right)}}}}\right| \right)} + \int{\sin{\left(x \right)} d x}$$

The integral of the sine is $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:

$$- a^{2} \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)} + {\color{red}{\int{\sin{\left(x \right)} d x}}} = - a^{2} \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)} + {\color{red}{\left(- \cos{\left(x \right)}\right)}}$$

Therefore,

$$\int{\left(- \frac{a^{2}}{\sin{\left(x \right)}} + \sin{\left(x \right)}\right)d x} = - a^{2} \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)} - \cos{\left(x \right)}$$

Add the constant of integration:

$$\int{\left(- \frac{a^{2}}{\sin{\left(x \right)}} + \sin{\left(x \right)}\right)d x} = - a^{2} \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)} - \cos{\left(x \right)}+C$$

$$$\int \left(- \frac{a^{2}}{\sin{\left(x \right)}} + \sin{\left(x \right)}\right)\, dx = \left(- a^{2} \ln\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right|\right) - \cos{\left(x \right)}\right) + C$$$A