<s>[INST] I integrated $\int_0^1 \sqrt{x-x^2} dx$ after a $u$-substitution and trig substitution, but I'm not sure what to do with the limits I'm a bit lost on the technique in this problem and could use some insight. When do we have to change the  limits?
$$\int_0^1 \sqrt{x-x^2} dx$$
completing the square gives me:
$$\int \sqrt{-x^2 +x - \frac{1}{4} + \frac{1}{4} } dx$$
$$\int \sqrt{-(x-\frac{1}{2})^2 + \frac{1}{4}} dx$$
$$ \int \sqrt{-u^2 + \frac{1}{4}} du$$
$$\int \sqrt{\frac{1}{4} - u^2} du$$'
Is trig sub the best way to go?
if $u = \frac{1}{2} \sin \theta$ then $du = \frac{1}{2} \cos \theta d \theta$
$$ = \int \sqrt{\frac{1}{4} - \frac{1}{4} \sin^2\theta} \frac{1}{2} cos \theta d \theta$$
$$ \int \frac{1}{2} cos \theta \frac{1}{2} cos \theta d \theta$$
$$\frac{1}{4} \int cos^2 \theta d \theta$$
using half angle identity I eventually get:
$$ \frac{1}{4} \int \frac{1}{2} ( 1 + cos(2 \theta)) d \theta$$
$$\frac{1}{8} \int (1 + cos(2 \theta) d \theta$$
$$ \frac{1}{8} ( \theta + \frac{1}{2} sin 2 \theta)$$
Where do I go from here? I'm a bit lost as to what to do with the u sub and how to get back to x from here.
 [/INST] Just so you are aware, this can be handled with the Beta and by extension the Gamma Function. 
Here you have:
\begin{align}
I &= \int_0^1 \sqrt{x - x^2}\:dx = \int_0^1 \sqrt{x\left(1  - x\right)}\:dx = \int_0^1 x^{\frac{1}{2}}\left(1  - x\right)^{\frac{1}{2}}\:dx\\& = B\left(\frac{1}{2} + 1, \frac{1}{2} + 1 \right)= B\left(\frac{3}{2} , \frac{3}{2}  \right)
\end{align}
Using the relationship between the Beta and the Gamma Function this becomes:
\begin{equation}
I = B\left(\frac{3}{2} , \frac{3}{2}  \right) = \frac{\Gamma\left(\frac{3}{2} \right)\cdot \Gamma\left(\frac{3}{2} \right)}{\Gamma\left(\frac{3}{2} + \frac{3}{2} \right)} = \frac{\Gamma\left(\frac{3}{2} \right)^2}{\Gamma\left(3\right)} = \frac{\left(\frac{\sqrt{\pi}}{2} \right)^2}{2} = \frac{\pi}{8}
\end{equation}
</s>