Integrals with limits I am trying to do:
$$\int_0^1 x\sin(180x^2)\, dx$$
I use substitution:
Let $ u = 180x^2 $ and $ \tfrac{du}{dx}=360x $
$$  \implies du =360x \,dx $$
$$  \implies \frac{1}{360} \, du = dx $$
so we get
$$ = 360\int_0^1 x\sin(u) \,du$$
$$\left[-\frac{1}{360}\cos(180x^2)+C\right]_0^1$$
$[-\tfrac{1}{360}\cos(180(1)^2)] \, - $ 
$[-\tfrac{1}{360}\cos(180(0)^2)] \, $ 
$$$$
$= \tfrac{1}{360} - \tfrac{1}{360} = 0$
However the answer on this site is different
http://www.integral-calculator.com/#expr=xsin%28180x%5E2%29&lbound=0&ubound=1
I dont understand how they came up with
$= \tfrac{1}{360} - \tfrac{\cos(180)}{360} = 0$
in the indefinite integral box, how did they get cos(180)?!
 A: You made a mistake going from $du = 360x\,dx$ to $\dfrac{1}{360}du = dx$. 
You should have gotten $\dfrac{1}{360}du = x\,dx$. Then, performing the substitution gives you 
$\displaystyle\int_{0}^{1}x\sin(180x^2)\,dx = \int_{0}^{180}\dfrac{1}{360}\sin u\,du = \left[-\dfrac{1}{360}\cos u\right]_{0}^{180}$ 
$= -\dfrac{1}{360}\cos(180)+\dfrac{1}{360}\cos(0) = \dfrac{1}{360} - \dfrac{1}{360}\cos(180)$. 
Note that by saying that the antiderivative of $\sin u$ is $-\cos u$, you are assuming the argument is in radians not degrees. The online calculator you are using also assumed the argument was in radians. Hence, $\cos(180) \neq -1$.
If your original problem was $\displaystyle\int_{0}^{1}x\sin(\pi x^2)\,dx$ and not $\displaystyle\int_{0}^{1}x\sin(180 x^2)\,dx$, then it would probably be less confusing if you left the argument in radians. The overall method will be the same. 
A: I'm going to answer in terms of radians because that is what the OP said that the problem was in. Note that a valid solution can be found using degrees, one must just be weary. $360$ without a unit often means $360 \mathrm{rad} \neq 360^{\circ}$. 
Note that $2 \pi =360^{\circ}$ and $\pi =180^{\circ}.$
For $\int x\sin(\pi x^2)\, dx$ let $u=\pi x^2$ and $du=2 \pi x$. Thus, we have $$\frac{1}{2 \pi} \int 2 \pi x\sin(\pi x^2)\, dx= \frac{1}{2 \pi} \int \sin(u)\, du= \frac{-1}{2\pi} \cos u = \frac{-1}{2\pi} \cos(\pi x^2).$$
To calculate the bounds: $$\int_0^1 x\sin(\pi x^2)\, dx= \frac{-1}{2\pi} \cos(\pi x^2)]_0^1=\frac{-1}{2 \pi} \cos(\pi(1)^2)-(\frac{-1}{2\pi} \cos(\pi(0)^2)= \frac{-1}{2 \pi}(-1-(1))=\frac{1}{\pi}$$
Overall, it seems that your problem is the difference between radians and degrees. I would strongly recommend doing all trigonometry based calculus in radians, as they are natural units to do computations in. Degrees based measurements can be extremely messy and cumbersome, because one must keep in mind that $\cos (360) \neq \cos (2\pi) = \cos {360^{\circ}}.$
