# $\int^{\frac{\pi }{4} }_{0} \cos ^{\frac{3}{2} }\left(2 \theta \right) \cos \left( \theta \right) d\theta$

$$\int^{\frac{\pi }{4} }_{0} \cos ^{\frac{3}{2} }\left(2 \theta \right) \cos \left( \theta \right) d\theta$$

This integration above I tried to solve it by and get

$$\int^{\frac{\pi }{4} }_{0} \left( 1-2\sin ^{2}\left( \theta \right) \right) ^{\frac{3}{2} }d\sin \left( \theta \right)$$

and I tried to evaluate the power but I find this is useless.

My question is: how I can get this integration in the closed form?
Thanks.

Define $\sin{\phi}=\sqrt{2} \sin{\theta}$. Then the integral is equal to

$$\frac1{\sqrt{2}} \int_0^{\pi/2} d\phi \, \cos^4{\phi} = \frac{3 \sqrt{2} \pi}{32}$$

NB

$$\int_0^{\pi/2} d\phi \, \cos^{2 n}{\phi} = \frac1{2^{2 n}} \binom{2 n}{n} \frac{\pi}{2}$$

• your solution gives approximately 0.265, when Wolfram Alpha returns an approximate value of 0.614. Edit: I can't tell what is the correct approximation. Anyone else have another software to check other than mathematica? – Brevan Ellefsen Sep 16 '15 at 20:31
• @BrevanEllefsen: I was off by a factor of 2 (forgot to multiply by $\pi/2$ rather than $\pi$). But Mathematica agrees with me. – Ron Gordon Sep 16 '15 at 20:34
• @BrevanEllefsen: would help if I put the $\pi$ in as well. Forgive me for my sloppiness. – Ron Gordon Sep 16 '15 at 20:35
• @BrevanEllefsen: the problem is for $3/2$. It is hard to see, but zoom in, or look at the script. – Ron Gordon Sep 16 '15 at 20:37
• @user257567: it is just a generalization of the result I obtained through the substitution, just so you know where the result comes from. My point was that we can write down the integrals of even powers of cosines over $[0,\pi/2]$, so that the substitution I provided reduced the integral to something well known. – Ron Gordon Sep 16 '15 at 20:46