How to find the antiderivative of $\sqrt{25-x^2}$?

Question

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.

Answer 1

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 $x$.

Answer 2

We often use antiderivatives to calculate area. For fun, let's use area to calculate an antiderivative.

We want to find a function $F(w)$ such that $F'(w)=\sqrt{25-w^2}$.
Let $F(w)$ be the area under the curve $y=\sqrt{25-x^2}$, above the $x$-axis, from $x=0$ to $x=w$. Then $F(w)$ is an antiderivative of $\sqrt{25-w^2}$.

We find this area $F(w)$. Draw the circle $x^2+y^2=w$. Let $O$ be the origin, let $W$ be the point $(w,0)$, and let $Y$ be the point $(0,5)$. Draw a vertical line through $W$, and suppose it meets the upper half of the circle at $P$.

Then the area $F(x)$ is the area of $\triangle OWP$ plus the area of the circular sector $OPY$.

It is easy to see that $\triangle OWP$ has area $$\frac{1}{2}w\sqrt{25-w^2}, \tag{$1$}$$ since it has base $w$ and height $\sqrt{25-w^2}$.

So now we only need to find the area of circular sector $OPY$. The angle of the sector is $\pi/2$ minus the angle whose cosine is $w/5$. To put it in more standard terms, the angle is $\arcsin(w/5)$. The radius of the circle is $5$, so the area of circular sector $OPY$ is $$\frac{1}{2}(5^2)\arcsin(w/5). \tag{$2$}$$

Finally, add $(1)$ and $(2)$ to find an antiderivative of $\sqrt{25-w^2}$.