Two variable function, single integral

[bananabandana]

Homework Statement

Evaluate:
I(y)= \int^{\frac{\pi}{2}}_{0} \frac{1}{y+cos(x)} \ dx if y > 1

Homework Equations

The Attempt at a Solution

I've never seen an integral like this before. I can see it has the form:
\int^{a}_{b} f(x,y) dx
I clearly can't treat it as one half of an exact differential of F(x,y) because \frac{\partial{u}^{2}}{\partial{x} \partial{y}} \neq \frac{\partial{u}^{2}}{\partial{y}\partial{x}}

I'm not sure if I can assume whether y is a constant or not. I think from the definition of integration it is possible just to say that y is independent of x. If that is so, I make the substitution u=y+cos(x) which leaves me with:
I(y) = \int^{y+1}_{y} \frac{1}{u} \times \frac{1}{sin(x)} du = \int^{y+1}_{y} \frac{1}{u \sqrt{1-(u-y)^{2}}} du

But this seems unintegrable.. which leaves me with a problem! Where have I gone wrong? Can I not assume y is a constant?

Thanks!

 

[SteamKing]

Since you are integrating w.r.t. x, you treat y as a constant.

 

[LCKurtz]

Perhaps there is an easier way of doing the definite integral, but normally for this type of integrand you need the tangent half angle substitution. See here for an example and explanation:
http://www.westga.edu/~faucette/research/Miracle.pdf

 

[bananabandana]

Thanks LCKurtz t=tan(\frac{x}{2}) works. Had completely forgotten about that, silly me :P