Question

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Answer 1

\[\int_0^{\pi/2}\int_0^y\frac{\cos2y}{\sqrt{9-4\sin^2x}}~dx~dy\]

Answer 2

Have you tried anything specific? Like changing order of integration, for instance

Answer 3

Okay, changing the order seems to me to be best thing to do here. The region over which the function \(\dfrac{\cos2y}{\sqrt{\cdots}}\) is being integrated is drawn below:

Where the region can be defined by
\[R_1:=\left\{(x,y)~:~0\le x\le y,~0\le y\le\frac{\pi}{2}\right\}\]
you can also describe it by
\[R_2:=\left\{(x,y)~:~0\le x\le \frac{\pi}{2},~0\le y\le x\right\}\]
so that the original integral is equivalent to
\[\int_0^{\pi/2}\int_0^x\frac{\cos2y}{\sqrt{9-4\sin^2x}}~dy~dx\]