1. ## double integral

Calculate the double integral where is the region:
I dun know how to integrate it. Do i have to integrate with x first or y first? even though i tried both way, no clue to integrate.

2. ok i think i figure out something..

$\int^{\pi/3}_{0}\int^{\pi/2}_{0} xcos(2x+y)dydx$
u= 2x+y
du = dy

$\int^{\pi/3}_{0}\int^{\pi/2}_{0} xcos(u)dudx$
$\int^{\pi/3}_{0} [xsin(u)]_{2x}^{2x+\pi/2}dx$
$\int^{\pi/3}_{0} [xsin(2x+ \pi/2) - xsin(2x)]dx$
$\int^{\pi/3}_{0} xsin(2x+ \pi/2) dx - \int^{\pi/3}_{0}xsin(2x)dx$

I'm stuck here. pls help

3. Originally Posted by zpwnchen
ok i think i figure out something..

$\int^{\pi/3}_{0}\int^{\pi/2}_{0} xcos(2x+y)dydx$
u= 2x+y
du = dy

$\int^{\pi/3}_{0}\int^{\pi/2}_{0} xcos(u)dudx$
$\int^{\pi/3}_{0} [xsin(u)]_{2x}^{2x+\pi/2}dx$
$\int^{\pi/3}_{0} [xsin(2x+ \pi/2) - xsin(2x)]dx$
$\int^{\pi/3}_{0} xsin(2x+ \pi/2) dx - \int^{\pi/3}_{0}xsin(2x)dx$

I'm stuck here. pls help
Now use integration by parts.

4. I did using by parts. But way too complicated.

u = x
du = dx

$dv = sin(2x+ \pi/2)dx$
$v = -1/2 cos(2x+ \pi/2)$

If i substitude them in, it become complicated. isnt it right?

5. Originally Posted by zpwnchen
I did using by parts. But way too complicated.

u = x
du = dx

$dv = sin(2x+ \pi/2)dx$
$v = -1/2 cos(2x+ \pi/2)$

If i substitude them in, it become complicated. isnt it right?
The whole point of integration by parts is that you end up with a simpler integral. In this case, you end up having to integrate $-\frac{1}{2} \cos \left(2x+ \frac{\pi}{2} \right)$ - surely that's not too difficult a job (especially if you're studying double integrals) ....?