How Do You Integrate cos²(π/2 cosθ) from 0 to π/2?

[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 !

 

[Millennial]

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

 

[arildno]

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

 

[micromass]

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 !

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?

 

[doey]

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?

(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

 

[micromass]

(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

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

 

[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..

 

[micromass]

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..

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.

 

[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.