Integral $\int{\sqrt{25 - x^2}dx}$ I'm trying to find $\int{\sqrt{25 - x^2} dx}$
Now I know that $\int{\frac{dx}{\sqrt{25 - x^2}}}$ would have been $\arcsin{\frac{x}{5}} + C$, but this integral I'm asking about has the rooted term in the numerator.
What are some techniques to evaluate this indefinite integral?

Since you already know that
$$\int{\frac{dx}{\sqrt{25 - x^2}}}=\arcsin{\frac{x}{5}} + C$$
you can actually skip the trigonometric substitution part and solve by partial integration:
$$\begin{array}{lcl}\int{\sqrt{25 - x^2} dx} & = & x\sqrt{25 - x^2} - \int{\frac{x (-2x)dx}{2\sqrt{25 - x^2}}} \\
& = & x\sqrt{25 - x^2} - \int{\frac{-x^2 dx}{\sqrt{25 - x^2}}} \\
& = & x\sqrt{25 - x^2} - \int{\frac{25-x^2 dx}{\sqrt{25 - x^2}}} + \int{\frac{25 dx}{\sqrt{25 - x^2}}} \\
& = & x\sqrt{25 - x^2} - \int{\sqrt{25 - x^2} dx} + 25\arcsin{\frac{x}{5}} + C \; .
\end{array}$$
Or after rearranging
$$\int{\sqrt{25 - x^2} dx} = \frac{1}{2} x\sqrt{25 - x^2} + \frac{25}{2}\arcsin{\frac{x}{5}} + C \; .$$
