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?

Oh man, this wasn't my idea, but I found it in a hint in Schaum's Calculus 5e,
If evaluated as a definite integral, on x=0 to x=5, can consider $ \int{ \sqrt{ 25 - x^2 } dx } $ as the area under a quarter circle of radius 5.
So
$ \int_0^5{ \sqrt{ 25 - x^2 } dx } $
$ = \frac{1}{4} \pi r^2 $
$ = \frac{ 25 \pi }{4} $
This won't work if you integrate only part of the circle however.
