<s>[INST] Double integral over $x^2+y^2 \le 1$ I am trying to calculate the double integral
$\displaystyle \iint_{x^2+y^2\leq 1} (\sin x+y+3)\,dA$
Here are my attempts so far:
1) I used polar coordinates 
$ x= r \sin(\theta)$
$y= r \cos (\theta)$ 
where  $\theta \in [0,2 \pi]$ and $r \in [0,1]$ which gives
$\displaystyle \int_0^{2\pi} \int_0^1 \bigg(\sin\big(r \cos(\theta)\big)+ r \sin(\theta)+3 \bigg)r\, dr\, d\theta$
and stuck with fiding antiderivative of the function with respect to r
$ r \sin\big(r \cos(\theta)\big)$
2) I tried to divide the region into parts that 
$A \cup B = \{ (x,y) : x^2+y^2 \leq 1\}$ where $A=\{ (x,y) : x^2+y^2 \leq 1$ and $x\geq 0 \}$ and $B=\{ (x,y) : x^2+y^2 \leq 1$ and $x <0 \}$ which gives me 
$ \quad\displaystyle \int_{-1}^1 \int _0^{\sqrt{1-x^2}} \big( \sin x +y+3 \big)\,dy \,dx + \int_{-1}^1 \int _0^{\sqrt{1-x^2}} \big( \sin x +y+3 \big)\,dy\,dx$
$\displaystyle =\int_{-1}^1 \big( \sqrt{1-x^2} \sin x + \frac{1-x^2}{2}+3 \sqrt{1-x^2} \big)\,dx + \dotsm $
and stuck with finding antiderivative of the function
$\sqrt{1-x^2} \sin x$.
I will be glad to hear any tips. Thanks in advance.
 [/INST] There is no reason not to split this into three separate integrals:
$$\iint_{x^2+y^2\leq 1} \sin x\;dA+\iint_{x^2+y^2\leq 1}y\;dA+\iint_{x^2+y^2\leq 1}3\;dA$$
You can do the third, I think. For the first, note that $\sin x$ is antisymmetric about the $y$-axis; and for the second, note that $y$ is antisymmetric about the $x$-axis. So there is nothing to evaluate!
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