How to find the antiderivative of $\sqrt{25-x^2}$? How to find the antiderivative of $\sqrt{25-x^2}$?
How would I do it? Integration by substitution doesn't seem to work in this case.

A trigonometric substitution can do the trick. Note that $1-\sin^2\theta = \cos^2\theta$.
Let $x = 5\sin\theta$, with $-\frac{\pi}{2}\leq \theta\leq\frac{\pi}{2}$. Then
$$\sqrt{25-x^2} = \sqrt{25-25\sin^2\theta} = 5\sqrt{\cos^2\theta} = 5|\cos\theta| = 5\cos\theta$$
(with the last equality because with $-\frac{\pi}{2}\leq \theta\leq\frac{\pi}{2}$, we have $\cos\theta\geq 0$.
We also have $dx = 5\cos\theta d\theta$. So
$$\begin{align*}
\int\sqrt{25-x^2}\,dx &= \int5\cos\theta 5\cos\theta\,d\theta\\
&= 25\int\cos^2\theta\,d\theta.
\end{align*}$$
The integral of $\cos^2\theta$ can be done by using Integration by Parts or a reduction formula. Then convert back to $x$.
