# Integration cos^2(∏/2cosθ)

1. Dec 3, 2012

### doey

i facing a maths problem in integrating ∫ cos^2(∏/2cosθ) with limit from 0 to ∏/2,i was panic and struggled a long period of time in solving this,anyone can help me? pls give me the answer in detail tq !

2. Dec 3, 2012

### Millennial

Use the integral identity $\displaystyle \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$.

3. Dec 3, 2012

### arildno

Why would that help, Millenial?
You change the internal cos(theta) to a sin(theta)..

4. Dec 3, 2012

### micromass

Staff Emeritus
First of all, do you mean

(1) $\int_0^{\pi/2} \cos^2 (\frac{\pi}{2} \cos\theta) d\theta$

or

(2) $\int_0^{\pi/2} \cos^2 (\frac{\pi}{2\cos \theta}) d\theta$

Also, which methods do you have at your disposal? Contour integration? Differentiation under the integral sign? Just normal calc II techniques?

5. Dec 3, 2012

### doey

(1) $\int_0^{\pi/2} \cos^2 (\frac{\pi}{2} \cos\theta) d\theta$
,i am asking this pls let me know the steps it takes

6. Dec 3, 2012

### micromass

Staff Emeritus
Which course is this for? Do you know Bessel functions?? The solution requires this (at least that is what wolfram alpha says).

7. Dec 3, 2012

### arildno

Just a thought:
We may easily rewrite this equation into the identity:
$$\int_{0}^{\frac{\pi}{2}}\cos^{2}(\frac{\pi}{2}\cos\theta)d\theta+\int_{0}^{\frac{\pi}{2}}\sin^{2}( \frac{\pi}{2}\cos\theta)d\theta=\frac{\pi}{2}$$
I feel dreadfully tempted to declare the two integrals to have the same value (the latter being merely a flipped version of the first), but temptation is not proof..

Last edited by a moderator: Dec 3, 2012
8. Dec 3, 2012

### micromass

Staff Emeritus
Hmmm, looking at the graph doesn't really convince me that the integrals are equal

Anyway, wolfram alpha gives us

$\int_0^{\pi/2} \cos^2(\frac{\pi}{2} \cos(x))dx = \frac{\pi}{4}(1+J_0(\pi))$

so I doubt the integral will be solvable with methods like these.

Last edited by a moderator: Dec 3, 2012
9. Dec 3, 2012

### Mute

To evaluate the integral one will have to use the identities

$$\cos t = \frac{1}{2}(e^{it}+e^{-it})$$
(or just $\cos t = \mbox{Re}[\exp(it)]$)

and

$$e^{iz\cos\theta} = \sum_{n=-\infty}^\infty i^n J_n(z)e^{in\theta}.$$

I guess the trig identity

$$\cos^2 t = \frac{1}{2}(1+\cos(2t))$$
also helps.

Last edited: Dec 3, 2012