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.

  • 1
    $\begingroup$ At first let $u=\cot x$ $\endgroup$ – Nosrati Jan 8 '17 at 21:14

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

| cite | improve this answer | |
  • $\begingroup$ Well, $1+\cot^2x=\csc^2x$ turns out to be an extremely useful identity. Thank you. $\endgroup$ – Glycerius Jan 8 '17 at 21:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.