Integrating with Substitution: Solving Tricky Integrals

[jamesbob]

Am i being really dumb when struggling to do this?

L = 4\sqrt{2}c \left \int_{0}^{\frac{\pi}{2}} \left \frac{dt}{\sqrt{1 + sin^2 t}}

Using substitution or otherwise show that

L = 4c \left \int_{0}^{1} \left \frac{du}{\sqrt{1 - u^4}}

Its a small part of a question but its stopping me doing the rest. Anyone help me out? The limits I am fine with, for the rest i get:

u = \sin t \left \frac{du}{dt} = \cos t \left dt = \frac{du}{\cos t} \left so \left L = \frac{du}{\sqrt{1 - u^2}\cos t} \left ?

 

[TD]

I think it should be a \sqrt {1 + u^2 } in the denominator, in what you have so far. Then remember that everything has to stand in the new variable u, you can't have any t's anymore.
But since u = \sin t, we have that \cos t = \sqrt {1 - \sin ^2 t} = \sqrt {1 - u^2 }.

Can you finish it?

 

[dextercioby]

BTW,

\int \frac{dt}{\sqrt{1+\sin^{2}t}} =F\left(x|-1\right)

,where F(x|m) is http://documents.wolfram.com/mathematica/functions/EllipticF/" .

Daniel.