Integrating $\sin(y^2)$ I've been stuck on this for over a half hour now and its quite frustrating.
$$\int_{0}^{9} \frac {y\sin(y^2)}{2}\,dy$$
Double angle formula is not usable here because were dealing with $\sin(y^2)$ and not $\sin^2(2y)$ or something like that. I hate being this type of question-poster, but I truly do not know where to start. I tried parts, letting 
$u=\frac{y}{2}$ and $dv=\sin(y^2)$ but I'm stuck at integrating $\sin(y^2)$ and would really like some help on this one.
 A: Hint
$$
I=\int_{0}^{9} \frac {y\sin(y^2)}{2}\,dy=\frac14\int_0^92y\sin(y^2)\,dy.
$$ 
I suppose that the $u$ substitution becomes quite clear.
I am sure that you can take from here.
A: Hint: make $u=y^2$ then $du=2y\,dy$
A: $$\int_{0}^{9} \frac {y\sin(y^2)}{2}\,dy = \int_{0}^{9} \sin(y^2)\,dy^2 =-\cos (y^2)\big|_{0}^9=1-\cos 81$$
A: Everyone is spot on with using substitution.
Not everyone understands how to spot a substitution problem though.  The idea behind using substitution is to take an integral that looks like Usain Bolt and turning it into the kid that gets picked last in gym class.  In our case we see $\,sin(y^{2})\,$ and think "woah, clam down satan".  However, we also notice that there is a $y$ outside of the $sin$ expression.  The light bulb inside us all lights up and we realize that we can use substitution.  If we let $u = y^{2}$ the integral becomes
$$ \int_{0}^{9} \frac{y \sin(y^{2})}{2}dy \,=\, \int_{0}^{9} \frac{y \sin(u)}{2}dy $$
We also have to make sure the bounds of our integration change so using the original inputs and our new inputs we have $u = y^{2} \,\Rightarrow\, u = 0^{2} \,=\, 0$ and $u = y^{2} \,\Rightarrow\, u = 9^{2} \,= \,81$. This leaves us with the integral 
$$ \int_{0}^{9} \frac{y \sin(u)}{2}dy \, = \, \int_{0}^{81} \frac{y \sin(u)}{2}dy$$
Unfortunately, our substitution isn't complete.  We still have that kid that refuses to put deodorant on after gym, or in our case: $dy$.  Looking at our original substitution we have $u = y^{2}$.  If we take the derivative of this equation we have
$$ u = y^{2} \,\Rightarrow \, \frac{du}{dy} = 2y $$
Hey look, there is a $dy$ in there.  Let's go ahead and acknowledge the fact that our deodorant-less class mate is stinky and separate him from the group
$$ \frac{du}{dy} = 2y \;\Rightarrow\; du = 2y \, dy \;\Rightarrow\; dy = \frac{du}{2y}$$
Fortunately for our nostrils our smelly friend, $dy$, is replaced by $du\,/\,2y$. Giving us
$$ \int_{0}^{81} \frac{y \sin(u)}{2}dy \, \Rightarrow \, \int_{0}^{81} \frac{y \sin(u)}{2} \frac{du}{2y}$$
Simplifying further we realize $y$ can suck it and get out of here. This leaves us with
$$ \int_{0}^{81} \frac{y \sin(u)}{2} \frac{du}{2y} \Rightarrow \int_{0}^{81} \frac{\sin(u)}{4}du$$
Lastly we can cordially invite the $\frac{1}{4}$ inside the integral to patiently wait until the adults inside the integral finish their business.  Leaving us with our final integration
$$ \frac{1}{4} \int_{0}^{81} \sin(u) \, du$$
