# Another couple of integration porblems

Back for a couple more I'm having a little trouble with. The first is

sech(1/x)tanh(1/x) all over x^2

Here I'm not sure how to deal with integrating hyperbolics at all as we haven't gotten to it in class yet. I did some searching on the internet and found that the integral tanh(x)sech(x) = -sech(x) + C. With all of it being over x^2 it becomes a little more complicated.

Do I rewrite x^2 as x^-2 and multiply it through with the 1/x's? Would the property tanh(x)sech(x) = -sech(x) + C still work?

I was also wondering about rewriting the whole equation as (2/e^1/x + e^-1/x)(e^1/x - e^-1/x /e^1/x - e^-1/x)(1/x^2). That seems overly complicated though and I'm not sure what form to go to from that point.

The other problem I'm having trouble with is integrating

cos^-3(2θ)sin(2θ)dθ

I know of ways of integrating sin and cos that have squares but is there a property I can use that's cubed and a double angel?

## Answers and Replies

Cyosis
Homework Helper
What do you mean by integrating all over x^2? I suppose you mean $$\int \text{sech}\left(\frac{1}{x}\right)\tanh\left(\frac{1}{x}\right)\frac{1}{x^2}dx$$. If so use the substitution $u=sech \left(\frac{1}{x}\right)$. For your second problem use the substitution $u=\cos 2\theta$

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Yeah, that is what I meant and both make so much more sense to me now. Thanks!