# 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.

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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 $x$.

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But why does that work and not Integration by Substitution? – manaraka May 30 '12 at 20:46
@manaraka: Not every integral can be done by simple substitution; that's why you have other techniques. Trigonometric substitution and integration by parts are some of those techniques. This integral cannot be done by a simple substitution for the simple reason that $\sqrt{25-x^2}$ is not the result of differentiating a simple composition. – Arturo Magidin May 30 '12 at 20:48
Why did you choose x=5sinθ? – manaraka May 30 '12 at 20:50
@manaraka: That's how trigonometric substitution works. When you are faced with a radical of the form $\sqrt{a^2-x^2}$, you substitute $x=a\sin\theta$ (or $x=a\cos\theta$) because that will simplify it; when faced with a radical of the form $\sqrt{a^2+x^2}$, you substitution $x=a\tan\theta$ because then $\sqrt{a^2+x^2} = \sqrt{a^2\sec^2\theta}$; when faced with a radical of the form $\sqrt{x^2-a^2}$, you use $x=a\sec\theta$. – Arturo Magidin May 30 '12 at 20:52
OK thanks, I get it now :D – manaraka May 30 '12 at 20:53

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 $\sqrt{25-w^2}$.

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