- #1
bananabandana
- 113
- 5
Homework Statement
Evaluate:
[tex] I(y)= \int^{\frac{\pi}{2}}_{0} \frac{1}{y+cos(x)} \ dx [/tex] if [itex] y > 1 [/itex]
Homework Equations
The Attempt at a Solution
I've never seen an integral like this before. I can see it has the form:
[itex] \int^{a}_{b} f(x,y) dx [/itex]
I clearly can't treat it as one half of an exact differential of [itex] F(x,y) [/itex] because [tex] \frac{\partial{u}^{2}}{\partial{x} \partial{y}} \neq \frac{\partial{u}^{2}}{\partial{y}\partial{x}} [/tex]
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 [itex] u=y+cos(x) [/itex] which leaves me with:
[tex] 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[/tex]
But this seems unintegrable.. which leaves me with a problem! Where have I gone wrong? Can I not assume y is a constant?
Thanks!