# In an MIT OOCW video the prof has the integral of $\frac{1}{(x^2+4)^3}$ where $x=2tan\,u$

In this video, the prof has used $$x=a\tan u$$ so $$x=2\tan u$$ for the integral of $$\cfrac{dx}{(x^2+4)^3}$$.

But I thought that could only be used with square roots of this case as in $$\displaystyle \int\cfrac{dx}{\sqrt{x^2+4}}$$

And without the square root then aren't you dealing with arctan?

• You can use any substitution you like, and if it helps you solve the integral, then so the better ! For example, in your case $x=2\tan u$ will be useful because when you square it you'll get $4\tan^2 u$ which means you can pull out a factor of $\frac{1}{4^3}$ and the integrand becomes $1/(1+\tan^2u)^3=1/\sec^6u=\cos^6u$. Bingo ! – Pixel Jul 18 at 17:39