Okay I have: $\displaystyle \int \frac {sec^{14}{20x}}{cot(20x)}$
I know I should break up $\displaystyle sec{x}$ but I am not sure what to do for the trig substitutions.
Thanks,
Matt
You're correct, you first need to split off a factor of $\displaystyle \sec(20x)$
$\displaystyle \int\frac{\sec^{14}(20x)}{\cot(20x)}\,dx=\int\frac {\sec^{13}(20x)\cdot\sec(20x)}{\cot(20x)}\,dx$
Now note that $\displaystyle \frac{1}{\cot(20x)}=\tan(20x)$
So we now see that $\displaystyle \int\frac{\sec^{13}(20x)\cdot\sec(20x)}{\cot(20x)} \,dx=\int \sec^{13}(20x)\cdot\sec(20x)\tan(20x)\,dx$
Now it should be clear what the proper substitution should be.
I hope this makes sense!
--Chris