How do you evaluate the integral of cotangent squared from Pi/4 to Pi/8?

[Briggs]

I have the question Evaluate $$\int_{Pi/4}^{Pi/8}_cot^2{2x}dx$$
So integrating this should (I hope) give $$[\frac{-1}{2}cot{2x}-x]$$ for those limits.
But I have never evaluated the cot integral before, I know that cotx=1/tanx. Do I substitute this identity in and work from there?

 

[HallsofIvy]

You could but I suspect that $\frac{1}{tan^2 x}$ would not be easy to integrate. How about just $cot(2x)= \frac{cos 2x}{sin 2x}$?

$$\int_{\pi/4}^{\pi/8}cot^2(2x)dx= \int_{\pi/4}^{\pi/8}\frac{cos^2(2x)}{sin^2(2x)}dx$$.

Since those are even powers you will need to use trig indentities to reduce them. By the way is there a reason for integrating from a larger value of x to a smaller?

 

[Briggs]

Ah that was a mistake, the x values should be the other way around.

So would $$\int_{\pi/8}^{\pi/4}cot^2(2x)dx= \int_{\pi/8}^{\pi/4}\frac{cos^2(2x)}{sin^2(2x)}dx$$ integrate to $$\frac{\frac{1}{2}sin^2(2x)}{\frac{-1}{2}cos^2(2x)}$$ ? Which I assume would simplify to $$-tan^2_(2x)$$

 

[TD]

Unfortunately not, but use the fact that $\cos ^2 \alpha + \sin ^2 \alpha = 1$ on:

$$\int {\frac{{\cos ^2 \left( {2x} \right)}}{{\sin ^2 \left( {2x} \right)}}dx} = \int {\frac{{1 - \sin ^2 \left( {2x} \right)}}{{\sin ^2 \left( {2x} \right)}}dx}$$

Then split the integral in two.