hello, everyone, I have been trying my best to solve this integral problem, but failed, can you help me?

Dec 2015
1,084
169
Earth
D∈ (-1/2,1/2) ; first only the integral , \(\displaystyle I=\int \dfrac{\sin^2 2x }{4\sqrt{4-(2x)^2 }}d(2x)=\int \dfrac{\sin^2 t-1+1 }{4\sqrt{4-t^2 }}dt=\int \dfrac{dt }{4\sqrt{4-t^2 }}-\int \dfrac{\cos^2 t }{4\sqrt{4-t^2 }}dt\).
$I=c+ \dfrac{\arcsin\left(\frac{x}{2}\right)}{4} -\displaystyle \int \dfrac{\cos^2 t }{4\sqrt{4-t^2 }}dt $ ; \(\displaystyle \; \) , the integral that remains is \(\displaystyle \int_D \dfrac{\cos^2 t }{4\sqrt{4-t^2 }}dt\). where t=2x.

Or better apply : \(\displaystyle \sin^2 t = 1/2 (1-\cos 2t)\).
 
Last edited:

romsek

Math Team
Sep 2015
2,964
1,674
USA
Mathematica throws up it's hands.
 
Jul 2008
5,248
58
Western Canada
Applying the identity, as idontknow did: cos(x)sin(x) = 1/2 sin(2x)
The closest I can find is in Gradshteyn and Ryzhik, page 435:
GRpg435a.png
Ho(a) is apparently the Struve function: Struve function - Wikipedia
Note that the limits of integration are different than those in the original problem, and the numerator is not squared.
But there was nothing else in the book, and it's a big book.

I don't suppose that this was intended to be a problem to be solved by numerical integration?
 
  • Like
Reactions: idontknow