Integral of $$$\csc^{3}{\left(x \right)}$$$
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Find $$$\int \csc^{3}{\left(x \right)}\, dx$$$.
Solution
For the integral $$$\int{\csc^{3}{\left(x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.
Let $$$\operatorname{u}=\csc{\left(x \right)}$$$ and $$$\operatorname{dv}=\csc^{2}{\left(x \right)} dx$$$.
Then $$$\operatorname{du}=\left(\csc{\left(x \right)}\right)^{\prime }dx=- \cot{\left(x \right)} \csc{\left(x \right)} dx$$$ (steps can be seen here) and $$$\operatorname{v}=\int{\csc^{2}{\left(x \right)} d x}=- \cot{\left(x \right)}$$$ (steps can be seen here).
The integral can be rewritten as
$$\int{\csc^{3}{\left(x \right)} d x}=\csc{\left(x \right)} \cdot \left(- \cot{\left(x \right)}\right)-\int{\left(- \cot{\left(x \right)}\right) \cdot \left(- \cot{\left(x \right)} \csc{\left(x \right)}\right) d x}=- \cot{\left(x \right)} \csc{\left(x \right)} - \int{\cot^{2}{\left(x \right)} \csc{\left(x \right)} d x}$$
Apply the formula $$$\cot^{2}{\left(x \right)} = \csc^{2}{\left(x \right)} - 1$$$:
$$- \cot{\left(x \right)} \csc{\left(x \right)} - \int{\cot^{2}{\left(x \right)} \csc{\left(x \right)} d x}=- \cot{\left(x \right)} \csc{\left(x \right)} - \int{\left(\csc^{2}{\left(x \right)} - 1\right) \csc{\left(x \right)} d x}$$
Expand:
$$- \cot{\left(x \right)} \csc{\left(x \right)} - \int{\left(\csc^{2}{\left(x \right)} - 1\right) \csc{\left(x \right)} d x}=- \cot{\left(x \right)} \csc{\left(x \right)} - \int{\left(\csc^{3}{\left(x \right)} - \csc{\left(x \right)}\right)d x}$$
The integral of a difference is the difference of integrals:
$$- \cot{\left(x \right)} \csc{\left(x \right)} - \int{\left(\csc^{3}{\left(x \right)} - \csc{\left(x \right)}\right)d x}=- \cot{\left(x \right)} \csc{\left(x \right)} + \int{\csc{\left(x \right)} d x} - \int{\csc^{3}{\left(x \right)} d x}$$
Thus, we get the following simple linear equation with respect to the integral:
$${\color{red}{\int{\csc^{3}{\left(x \right)} d x}}}=- \cot{\left(x \right)} \csc{\left(x \right)} + \int{\csc{\left(x \right)} d x} - {\color{red}{\int{\csc^{3}{\left(x \right)} d x}}}$$
Solving it, we obtain that
$$\int{\csc^{3}{\left(x \right)} d x}=- \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{\int{\csc{\left(x \right)} d x}}{2}$$
Rewrite the cosecant as $$$\csc\left(x\right)=\frac{1}{\sin\left(x\right)}$$$:
$$- \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{{\color{red}{\int{\csc{\left(x \right)} d x}}}}{2} = - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{\sin{\left(x \right)}} d x}}}}{2}$$
Rewrite the sine using the double angle formula $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:
$$- \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{\sin{\left(x \right)}} d x}}}}{2} = - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)}} d x}}}}{2}$$
Multiply the numerator and denominator by $$$\sec^2\left(\frac{x}{2} \right)$$$:
$$- \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} \right)} \cos{\left(\frac{x}{2} \right)}} d x}}}}{2} = - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2 \tan{\left(\frac{x}{2} \right)}} d x}}}}{2}$$
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 here), and we have that $$$\sec^{2}{\left(\frac{x}{2} \right)} dx = 2 du$$$.
So,
$$- \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2 \tan{\left(\frac{x}{2} \right)}} d x}}}}{2} = - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$
The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$
$$- \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$
Recall that $$$u=\tan{\left(\frac{x}{2} \right)}$$$:
$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} = \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} \right)}}}}\right| \right)}}{2} - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2}$$
Therefore,
$$\int{\csc^{3}{\left(x \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)}}{2} - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2}$$
Add the constant of integration:
$$\int{\csc^{3}{\left(x \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)}}{2} - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2}+C$$
Answer: $$$\int{\csc^{3}{\left(x \right)} d x}=\frac{\ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)}}{2} - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2}+C$$$