# Finding a definite integral

• May 8th 2010, 03:56 AM
maliky
Finding a definite integral
The question is:
Find the definite integral:
$\displaystyle \int\limits_0^{\pi/2} \frac{1}{1+cos(\alpha)cos(x)}dx$
($\displaystyle \alpha$ is a constant)

I've tried using the t-substitution, but I got quite a messy answer.
Possibly there's some simplification involving the fact that the constant is written in terms of cos that I'm not seeing?

The solution given is
Spoiler:
$\displaystyle \frac{\alpha}{\sin(\alpha)}$

Thanks.
• May 8th 2010, 04:55 AM
Prove It
Quote:

Originally Posted by maliky
The question is:
Find the definite integral:
$\displaystyle \int\limits_0^{\pi/2} \frac{1}{1+cos(\alpha)cos(x)}dx$
($\displaystyle \alpha$ is a constant)

I've tried using the t-substitution, but I got quite a messy answer.
Possibly there's some simplification involving the fact that the constant is written in terms of cos that I'm not seeing?

The solution given is
Spoiler:
$\displaystyle \frac{\alpha}{\sin(\alpha)}$

Thanks.

http://www.wolframalpha.com/input/?i...Cos&#91;x]%29]

Click Show Steps
• May 8th 2010, 06:32 AM
Random Variable
Universal substitution is probably the way to go.

$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos \alpha \cos x} \ dx$

let $\displaystyle t = \tan \Big(\frac{x}{2} \Big)$

then $\displaystyle dx = \frac{2 \ dt}{1+t^{2}}$

and $\displaystyle \cos x = \frac{1-t^{2}}{1+t^{2}}$

so $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos \alpha \cos x} \ dx = \int_{0}^{1} \frac{\frac{2}{1+t^{2}}}{1+ \cos \alpha \frac{1-t^{2}}{1+t^{2}}} \ dt$

$\displaystyle = \int^{1}_{0} \frac{2 \ dt}{1+\cos \alpha + t^{2} (1- \cos \alpha)} \ dt = \frac{2}{1+ \cos \alpha} \int^{1}_{0} \frac{dt}{1+ \Big(t \sqrt{\frac{1-\cos \alpha}{1+ \cos \alpha}}\Big)^{2}}$

$\displaystyle = \frac{2 \sqrt{1+\cos \alpha}}{(1+\cos \alpha)\sqrt{1-\cos \alpha}} \int_{0}^{\sqrt{\frac{1-\cos \alpha}{1+ \cos \alpha}}} \frac{du}{1+u^{2}}$

$\displaystyle = \frac{2}{\sqrt{1+\cos \alpha}\sqrt{1-\cos \alpha}} \arctan \Big( \sqrt{\frac{1-\cos \alpha}{1+ \cos \alpha}}\Big)$

$\displaystyle = \frac{2}{\sqrt{1-\cos^{2}\alpha}} \arctan \Big( \sqrt{\frac{1-\cos \alpha}{1+ \cos \alpha}}\Big) = \frac{2}{\sin \alpha } \arctan \Big( \sqrt{\frac{1-\cos \alpha}{1+ \cos \alpha}}\Big)$

now if $\displaystyle \alpha = \frac{\pi}{2}$, then we have $\displaystyle \arctan 1 = \frac{\pi}{4} = \frac{\alpha}{2}$

so $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cos \alpha \cos x} \ dx = \frac{2}{\sin \alpha} \frac{\alpha}{2} = \frac{\alpha}{\sin \alpha}$