# Integration substitution

1. Feb 28, 2006

### 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 im fine with, for the rest i get:

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

2. Mar 1, 2006

### 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?

3. Mar 1, 2006

### 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/" [Broken].

Daniel.

Last edited by a moderator: May 2, 2017