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

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?


1 Answer 1


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

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


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

  • $\begingroup$ 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? $\endgroup$ Sep 16, 2015 at 20:31
  • $\begingroup$ @BrevanEllefsen: I was off by a factor of 2 (forgot to multiply by $\pi/2$ rather than $\pi$). But Mathematica agrees with me. $\endgroup$
    – Ron Gordon
    Sep 16, 2015 at 20:34
  • $\begingroup$ @BrevanEllefsen: would help if I put the $\pi$ in as well. Forgive me for my sloppiness. $\endgroup$
    – Ron Gordon
    Sep 16, 2015 at 20:35
  • 1
    $\begingroup$ @BrevanEllefsen: the problem is for $3/2$. It is hard to see, but zoom in, or look at the script. $\endgroup$
    – Ron Gordon
    Sep 16, 2015 at 20:37
  • 1
    $\begingroup$ @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. $\endgroup$
    – Ron Gordon
    Sep 16, 2015 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.