Integral of $\csc^{3}{\left(x \right)}$

Find $\int \csc^{3}{\left(x \right)}\, dx$.

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 ») and $\operatorname{v}=\int{\csc^{2}{\left(x \right)} d x}=- \cot{\left(x \right)}$ (steps can be seen »).

The integral becomes

$$\int{\csc^{3}{\left(x \right)} d x}=\csc{\left(x \right)} \cdot \left(- \cot{\left(x \right)}\right)-\int{\left(- \cot{\left(x \right)}\right) \cdot \left(- \cot{\left(x \right)} \csc{\left(x \right)}\right) d x}=- \cot{\left(x \right)} \csc{\left(x \right)} - \int{\cot^{2}{\left(x \right)} \csc{\left(x \right)} d x}$$

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 »), and we have that $\sec^{2}{\left(\frac{x}{2} \right)} dx = 2 du$.

Therefore,

$$- \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2 \tan{\left(\frac{x}{2} \right)}} d x}}}}{2} = - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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:

$$\int{\csc^{3}{\left(x \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right| \right)}}{2} - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2}+C$$

$\int \csc^{3}{\left(x \right)}\, dx = \left(\frac{\ln\left(\left|{\tan{\left(\frac{x}{2} \right)}}\right|\right)}{2} - \frac{\cot{\left(x \right)} \csc{\left(x \right)}}{2}\right) + C$A