Integral $\int (\cot^{2}x+\cot^{4}x)\mathrm{d}x$

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I am having difficulties with integrating this function:$$\int (\cot^{2}x+\cot^{4}x)\mathrm{d}x$$I separated the function in two and calculated $\int \cot^2x\mathrm{d}x$:$$\int \cot^2x\mathrm{d}x=\int (-1+1+\cot^2x)\mathrm{d}x=-\int \mathrm{d}x+\int (1+\cot^2x)\mathrm{d}x=-x-\cot x+c$$But I don't know how to deal with $\int \cot^4x\mathrm{d}x$. I assume that there is a shorter solution somewhere but I'm not sure. In addition, this is a high school level question, I'm only familiar with primitives of simple functions, integration by parts and trigonometric replacement.

Any help would be appreciated, thanks in advance.

Solution$$\int (\cot^{2}x+\cot^{4}x)\mathrm{d}x=\int \cot^{2}x\cdot(\cot^{2}x+1)\mathrm{d}x=\int (\cot^{2}x\cdot\csc^{2}x)\mathrm{d}x$$Let's assume $u:=\cot x$

Then $\mathrm{d}u=-\csc^2x\mathrm{d}x$. So we have:$$\int (\cot^{2}x\cdot\csc^{2}x)\mathrm{d}x=-\int u^2\mathrm{d}u=-\frac{u^3}{3}+c=-\frac{\cot^2x}{3}+c$$

N. S. and MyGlasses, thank you for your help.

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1 Answer

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$$\int (\cot^{2}x+\cot^{4}x)\mathrm{d}x= \int\cot^{2}x\left(1+\cot^{2}x \right)\mathrm{d}x= \int\cot^{2}x\csc^{2}x \mathrm{d}x$$ and use the fact that $(\cot(x))'=-\csc^2(x)$.

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