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.

The integral of a sum is equal to the sum of the integrals of the summands (assuming all exist, and finitely many summands).
$$\iint_{x^2+y^2=1}\sin(x)+y+3\,dA = \iint_{x^2+y^2=1}\sin(x)\,dA + \iint_{x^2+y^2=1}y\,dA + \iint_{x^2+y^2=1}3\,dA$$
Hint: For any odd function $f$ (a function such that $f(-x) = -f(x)$ for all $x$), what is $\int_{-c}^cf(x)dx$?
