Here the equation: $\iint_{A} x^2y \ cos(xy^2)dxdy \ \ | \ \ A = <0, \frac\pi 2> x <0,2> $
Now I'm aware that it means integration on a rectangle with boundaries defined by given coordinates, which means that we should be able to rewrite it like this:
$\int^{\frac \pi 2}_{0} [\ \int^{2}_{0} x^2y \ cos(xy^2)dy \ ]dx$ .
From what I know, I guess it has to be calculated using integration by parts, however, but I have troubles defining u and v' because of both the variables being located in the argument of cosine. Also substitution has come to my mind, but I'm not sure what to substitute, since the powers in the argument of cosine and the multiplier do not correspond to each other.
I'd love at least some general advice as how to calculate this. According to the results page and Wolfram Alpha, it is equal to $ \frac \pi {-16} $
PS. I'm sorry, it's probably trivial, but I'm quite new into this multiple integrals thing..
Thanks in advance
This integral can be solved by substitution. In fact the needed substitution does not mix $x$ and $y$, so we only need to apply single variable calculus.
The needed substitution is $t=y^2$, because then $2y\, dy=dt$ which simplifies the integral a lot.
So lets first solve the $y$-integral
$$\int_0^2 x^2 y \cos(x y^2) \, dy= \frac{1}{2}\int_0^4 x^2 \cos(xt)dt = \frac{1}{2} x \sin(4x)$$
The value of the double integral is then obtained by integrating this expression with respect to $x \in [0,\frac{\pi}{2}]$.
This integral can be calculated by integration by parts: $$ \frac{1}{2}\int_0^{\pi/2} x \sin(4x)= \frac{1}{2} \left(\big[-\frac{x}4 \cos(4x)\big]_0^{\pi/2} +\int_0^{\pi/2}\frac{1}{4} \cos(4x)\right)$$ As the integral on the right is over a whole period of $\cos(4x)$, it vanishes. Thus the final result is $$-\frac{\pi}{16}.$$