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

#### idontknow

anyone knows where to start ?

#### idontknow

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 : $$\displaystyle \sin^2 t = 1/2 (1-\cos 2t)$$.

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#### romsek

Math Team
Mathematica throws up it's hands.

#### Yooklid

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:

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?

idontknow