Recent content by Maburo

After a few more substitutions, a few algebra tricks, and a lot of simplification, I think I have gotten the answer. It required only change of variables. \int\frac{dx}{\sqrt{1-\sin^4(x)}} = \frac{\sqrt{2}}{4}\ln\Biggr\vert 4\tan^2(x)+2\tan(x)\sqrt{2+4\tan^2(x)}+1 \Biggr\vert + C I was pretty...

If I am understanding correctly, you mean to write \int\frac{dx}{\sqrt{1-\sin^4(x)}} = \int\frac{dx}{\cos^2(x)\sqrt{1+2\tan^2(x)}} An obvious substitution for this would then be u=tan(x) \implies \int\frac{dx}{\cos^2(x)\sqrt{1+2\tan^2(x)}} = \int\frac{du}{\sqrt{1+2u^2}} This is much easier to...

Homework Statement Evaluate the following integral using a change of variables: \int\frac{dx}{\sqrt{1-\sin^4{x}}} Homework Equations If f(x)=g(u(x))u'(x) and \int g(x)dx = G(x) +C then \int f(x)dx = G(u(x))+C The Attempt at a Solution It seems helpful to first simplify a little to obtain...

Hello everyone. I am new here from Canada. I am looking forward to getting to know the forum!