Integral of $$$\cot^{4}{\left(3 x \right)}$$$

The calculator will find the integral/antiderivative of $$$\cot^{4}{\left(3 x \right)}$$$, with steps shown.

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Find $$$\int \cot^{4}{\left(3 x \right)}\, dx$$$.

Solution

Let $$$u=3 x$$$.

Then $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.

Thus,

$${\color{red}{\int{\cot^{4}{\left(3 x \right)} d x}}} = {\color{red}{\int{\frac{\cot^{4}{\left(u \right)}}{3} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(u \right)} = \cot^{4}{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\cot^{4}{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\int{\cot^{4}{\left(u \right)} d u}}{3}\right)}}$$

Let $$$v=\cot{\left(u \right)}$$$.

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

So,

$$\frac{{\color{red}{\int{\cot^{4}{\left(u \right)} d u}}}}{3} = \frac{{\color{red}{\int{\left(- \frac{v^{4}}{v^{2} + 1}\right)d v}}}}{3}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=-1$$$ and $$$f{\left(v \right)} = \frac{v^{4}}{v^{2} + 1}$$$:

$$\frac{{\color{red}{\int{\left(- \frac{v^{4}}{v^{2} + 1}\right)d v}}}}{3} = \frac{{\color{red}{\left(- \int{\frac{v^{4}}{v^{2} + 1} d v}\right)}}}{3}$$

Since the degree of the numerator is not less than the degree of the denominator, perform polynomial long division (steps can be seen »):

$$- \frac{{\color{red}{\int{\frac{v^{4}}{v^{2} + 1} d v}}}}{3} = - \frac{{\color{red}{\int{\left(v^{2} - 1 + \frac{1}{v^{2} + 1}\right)d v}}}}{3}$$

Integrate term by term:

$$- \frac{{\color{red}{\int{\left(v^{2} - 1 + \frac{1}{v^{2} + 1}\right)d v}}}}{3} = - \frac{{\color{red}{\left(- \int{1 d v} + \int{v^{2} d v} + \int{\frac{1}{v^{2} + 1} d v}\right)}}}{3}$$

Apply the constant rule $$$\int c\, dv = c v$$$ with $$$c=1$$$:

$$- \frac{\int{v^{2} d v}}{3} - \frac{\int{\frac{1}{v^{2} + 1} d v}}{3} + \frac{{\color{red}{\int{1 d v}}}}{3} = - \frac{\int{v^{2} d v}}{3} - \frac{\int{\frac{1}{v^{2} + 1} d v}}{3} + \frac{{\color{red}{v}}}{3}$$

Apply the power rule $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

$$\frac{v}{3} - \frac{\int{\frac{1}{v^{2} + 1} d v}}{3} - \frac{{\color{red}{\int{v^{2} d v}}}}{3}=\frac{v}{3} - \frac{\int{\frac{1}{v^{2} + 1} d v}}{3} - \frac{{\color{red}{\frac{v^{1 + 2}}{1 + 2}}}}{3}=\frac{v}{3} - \frac{\int{\frac{1}{v^{2} + 1} d v}}{3} - \frac{{\color{red}{\left(\frac{v^{3}}{3}\right)}}}{3}$$

The integral of $$$\frac{1}{v^{2} + 1}$$$ is $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:

$$- \frac{v^{3}}{9} + \frac{v}{3} - \frac{{\color{red}{\int{\frac{1}{v^{2} + 1} d v}}}}{3} = - \frac{v^{3}}{9} + \frac{v}{3} - \frac{{\color{red}{\operatorname{atan}{\left(v \right)}}}}{3}$$

Recall that $$$v=\cot{\left(u \right)}$$$:

$$- \frac{\operatorname{atan}{\left({\color{red}{v}} \right)}}{3} + \frac{{\color{red}{v}}}{3} - \frac{{\color{red}{v}}^{3}}{9} = - \frac{\operatorname{atan}{\left({\color{red}{\cot{\left(u \right)}}} \right)}}{3} + \frac{{\color{red}{\cot{\left(u \right)}}}}{3} - \frac{{\color{red}{\cot{\left(u \right)}}}^{3}}{9}$$

Recall that $$$u=3 x$$$:

$$\frac{\cot{\left({\color{red}{u}} \right)}}{3} - \frac{\cot^{3}{\left({\color{red}{u}} \right)}}{9} - \frac{\operatorname{atan}{\left(\cot{\left({\color{red}{u}} \right)} \right)}}{3} = \frac{\cot{\left({\color{red}{\left(3 x\right)}} \right)}}{3} - \frac{\cot^{3}{\left({\color{red}{\left(3 x\right)}} \right)}}{9} - \frac{\operatorname{atan}{\left(\cot{\left({\color{red}{\left(3 x\right)}} \right)} \right)}}{3}$$

Therefore,

$$\int{\cot^{4}{\left(3 x \right)} d x} = - \frac{\cot^{3}{\left(3 x \right)}}{9} + \frac{\cot{\left(3 x \right)}}{3} - \frac{\operatorname{atan}{\left(\cot{\left(3 x \right)} \right)}}{3}$$

Add the constant of integration:

$$\int{\cot^{4}{\left(3 x \right)} d x} = - \frac{\cot^{3}{\left(3 x \right)}}{9} + \frac{\cot{\left(3 x \right)}}{3} - \frac{\operatorname{atan}{\left(\cot{\left(3 x \right)} \right)}}{3}+C$$

Answer

$$$\int \cot^{4}{\left(3 x \right)}\, dx = \left(- \frac{\cot^{3}{\left(3 x \right)}}{9} + \frac{\cot{\left(3 x \right)}}{3} - \frac{\operatorname{atan}{\left(\cot{\left(3 x \right)} \right)}}{3}\right) + C$$$A


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