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?
 A: The technique is sometimes called "trigonometric substitution" and is often used when you have something of the form $\sqrt{a^2-x^2}$, $\sqrt{x^2-a^2}$, or $\sqrt{x^2+a^2}$ (though the specific substitutions are different in each case).  Suppose you have a right triangle with sides of length $a$, $x$, and whichever of the three square-root expressions above appears in your integral.  Label one of the non-right angles of the triangle as $\theta$.  Express $x$ in terms of $\theta$ so that you can express $dx$ in terms of $\theta$ and $d\theta$. and carry out a substitution in your integral to rewrite the integral in terms of $\theta$.  Hopefully, this results in an integral that is easier to evaluate.  After evaluating the indefinite integral in $\theta$, you probably want to substitute back to get an answer in terms of $x$.
In your particular integral, $\sqrt{25-x^2}=\sqrt{5^2-x^2}$, so $a=5$.  The right triangle to consider has hypotenuse 5, one leg of length $x$ and $\sqrt{25-x^2}$ is the length of the other leg.  If you let $\theta$ be the measure of the acute angle opposite $x$, then $\sin\theta=\frac{x}{5}$ or $x=5\sin\theta$ and $\frac{\sqrt{25-x^2}}{5}=\cos\theta$.  I'll leave the rest of the substitution and work to you.
A: 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.
A: 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
$$\int{\sqrt{25 - x^2} dx} = \frac{1}{2} x\sqrt{25 - x^2} + \frac{25}{2}\arcsin{\frac{x}{5}} + C \; .$$
A: substitution $\sqrt{25-x^2}=t(5-x)$
