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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?

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Read up about change of variables.You want to let $x=5 \sin u$ –  picakhu Mar 26 '11 at 18:38
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5 Answers 5

up vote 11 down vote accepted

There is a basic approach here, called "trigonometric substitution." This is something to use whenever you have an integral that contains a radical of any of the following forms: $$\sqrt{a^2-x^2},\qquad \sqrt{a^2+x^2},\qquad \sqrt{x^2-a^2},\qquad a\gt 0.$$ The way trigonometric substitution works is by using the trigonometric identities $$\sin^2\theta + \cos^2\theta = 1,\qquad \tan^2\theta + 1 = \sec^2\theta.$$ The idea is to replace $x$ with something that will turn the stuff inside the square root into a square, so that you can then eliminate the radical.

Here, with $\sqrt{25-x^2}$, the idea is to use the identity $1-\sin^2\theta=\cos^2\theta$. If we replace $x$ with something that will turn $25-x^2$ into, say, $25(1-\sin^2\theta)$, then when we take the square root things will simplify.

So, set $x=5\sin\theta$. Why $5$? Because we have $\sqrt{5^2 - x^2}$.

Notice also that $-5\leq x\leq 5$ must hold for the radical to make sense. That's good, because $5\sin\theta$ can only take values there. To make this substitution reversible, we must make the assignment one-to-one (a single value of $\theta$ for each value of $x$); the easiest way to do this is to restrict ourselves to $\theta$ in the range of the $\arcsin$. So to be complete, our substitution will be $$ x = 5\sin\theta,\qquad -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}.$$ Now, what happens if we do this? Well, for the radical, we have: $$\sqrt{25 - x^2} = \sqrt{25 - 25\sin^2\theta} = \sqrt{25(1-\sin^2\theta)} = \sqrt{25\cos^2\theta} = \sqrt{(5\cos\theta)^2} = |5\cos\theta|.$$ (Don't forget the absolute values! In general, $\sqrt{a^2}=|a|$). But... because $\theta$ is supposed to be between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, then $\cos\theta\geq 0$. So $|5\cos\theta| = 5\cos\theta$, without the absolute value. So this is great! The radical completely simplifies to $$\sqrt{25-x^2} = 5\cos\theta.$$ The other bit we need to compute is $dx$, since we are doing a change of variable. Since $x=5\sin\theta$, then $dx = 5\cos\theta\,d\theta$. So in summary, we have: $$\int\sqrt{25-x^2}\,dx = \int (5\cos\theta)(5\cos\theta)\,d\theta = \int 25\cos^2\theta\,d\theta.$$ So now we need to do the integral of $\cos^2\theta$. This is a standard integral (or can be done with Weierstrass $t$-substutition to get a rational function). It can be done by using integration by parts and then solving: taking $u=\cos\theta$, $dv=\cos\theta\,d\theta$, we get: $$\begin{align*} \int\cos^2\theta\,d\theta &= \cos\theta\sin\theta +\int\sin^2\theta\,d\theta\\ &= \cos\theta\sin\theta + \int(1-\cos^2\theta)\,d\theta\\ &= \cos\theta\sin\theta + \theta - \int\cos^2\theta\,d\theta. \end{align*}$$ Solving for $\int\cos^2\theta\,d\theta$, we get $$\int\cos^2\theta\,d\theta = \frac{1}{2}\cos\theta\sin\theta + \frac{1}{2}\theta + C.$$ Going back to the original integral, we have: $$\begin{align*} \int\sqrt{25-x^2}\,dx &= \int 25\cos^2\theta\,d\theta\\ &= 25\left(\frac{1}{2}\cos\theta\sin\theta + \frac{1}{2}\theta\right) + C\\ &= \frac{25}{2}\cos\theta\sin\theta + \frac{25}{2}\theta + C. \end{align*}$$ Finally, we need to "go back to $x$". Since $x=5\sin\theta$, then $\sin\theta = \frac{1}{5}x$, $\cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1 - \frac{x^2}{25}} = \frac{1}{5}\sqrt{25-x^2}$; and $\theta = \arcsin\left(\frac{x}{5}\right)$. So we have: $$\begin{align*} \int\sqrt{25-x^2}\,dx &= \frac{25}{2}\cos\theta\sin\theta + \frac{25}{2}\theta + C\\ &= \frac{25}{2}\left(\frac{1}{5}\sqrt{25-x^2}\right)\left(\frac{x}{5}\right) + \frac{25}{2}\arcsin\left(\frac{x}{5}\right) + C\\ &= \frac{1}{2}x\sqrt{25-x^2} + \frac{25}{2}\arcsin\frac{x}{5} + C. \end{align*}$$ You can verify this by differentiation: $$\begin{align*} &\frac{d}{dx}\left(\frac{1}{2}x\sqrt{25-x^2} + \frac{25}{2}\arcsin\frac{x}{5}+C\right)\\ &=\frac{1}{2}\sqrt{25-x^2}-\frac{1}{2}\frac{x^2}{\sqrt{25-x^2}} + \frac{25}{2}\left(\frac{1}{\sqrt{ 1 - \left(\frac{x}{5}\right)^2}}\right)\frac{1}{5}\\ &= \frac{1}{2}\sqrt{25-x^2} - \frac{1}{2}\frac{x^2}{\sqrt{25-x^2}} + \frac{5}{2}\left(\frac{1}{\frac{1}{5}\sqrt{25-x^2}}\right)\\ &= \frac{1}{2}\sqrt{25-x^2} - \frac{1}{2}\frac{x^2}{\sqrt{25-x^2}} + \frac{25}{2}\frac{1}{\sqrt{25-x^2}}\\ &= \frac{1}{2}\sqrt{25-x^2} + \frac{1}{2}\left(\frac{25 - x^2}{\sqrt{25-x^2}}\right)\\ &= \frac{1}{2}\sqrt{25-x^2} + \frac{1}{2}\sqrt{25-x^2} = \sqrt{25-x^2}. \end{align*}$$

In general:

  • If you have a radical $\sqrt{a^2-x^2}$, you can try a substitution $x = a\sin\theta$, $-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}$, which will transform $\sqrt{a^2-x^2}$ into $a\cos\theta$.

  • If you have a radical $\sqrt{a^2+x^2}$, you can try a substitution $x=a\tan\theta$, with $-\frac{\pi}{2}\lt x\lt \frac{\pi}{2}$, which will transform $\sqrt{a^2+x^2}$ into $a\sec\theta$.

  • If you have a radical $\sqrt{x^2-a^2}$, you can try a substitution $x = a\sec\theta$ with $-\frac{\pi}{2}\lt x \lt \frac{\pi}{2}$, which will transform $\sqrt{x^2-a^2}$ into $a\tan\theta$.

Once you do the substitution, you'll have an integral involving sines and cosines; these can be solved, in the worst case scenario, by using Weierstrass's $t$-substitution to change them into integrals of rational functions and doing partial fractions; and sometimes just directly.

There are other alternatives: you can use hyperbolic functions, for example, which satisfy $$\cosh^2\theta - \sinh^2\theta = 1,\qquad 1 - \tanh^2\theta = \frac{1}{\cosh^2\theta},$$ so you can use "hyperbolic substitution" instead of trigonometric substitution. The result may need further transformation (just as in the trigonometric case).

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+1 very well detailed answer. –  fdart17 Mar 26 '11 at 21:11
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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.

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For the interested reader, you can also do hyperbolic trig substitution with these problems. It's similar but you'd need to learn all the hyperbolic trig identities and such so I stick with regular trig substitution myself. –  Graphth Mar 26 '11 at 19:43
    
Wow. If I had read this article on trig sub before "learning" it in class, I would have been much better off! Well done. +.99999... –  The Chaz 2.0 Mar 26 '11 at 20:56
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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.

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This is actually what comes to mind first when I see an integral like yours, but yours was indefinite, so I didn't even mention it. For definite integrals, it's worth keeping in mind that they are a measure of area and sometimes the geometric interpretation of the area is much simpler than the calculus interpretation. –  Isaac Mar 26 '11 at 19:16
    
This is the first thing that came to my mind too. –  Adrian Petrescu Mar 26 '11 at 21:07
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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

$$\int{\sqrt{25 - x^2} dx} = \frac{1}{2} x\sqrt{25 - x^2} + \frac{25}{2}\arcsin{\frac{x}{5}} + C \; .$$

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substitution $\sqrt{25-x^2}=t(5-x)$

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