Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers 2

up vote 5 down vote accepted

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 $x$.

share|improve this answer
1  
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
3  
@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
show 1 more comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.