# Troubleshooting: Evaluating an Trigonometric Integral Algebraically

Hehe…we are always amused by calculus students who manage to mis-integrate a function to the point of yielding nonsensical answers. To illustrate what we mean by this, here is Steve, who — at the point of this writing at least — still haven’t got a clue where his computations went awry:

###### How do you evaluate $\displaystyle \int_{-5}^5 x – \sqrt{25-x^2} \, dx$ algebraically? When I take the areas geometrically, I get $\displaystyle -\frac{25\pi}{2}$, as $\displaystyle \int_{-5}^5 x \, dx=0$ and $\displaystyle \int_{-5}^5 \sqrt{25-x^2} \, dx$ is the area of a half-circle with radius $5$. But when I try to evaluate it algebraically, I don’t seem to get the same answer: $\displaystyle \int^5_{-5}\left(x-\sqrt{25-x^2}\right)\,dx=\left[\frac{x^2}{2}-\frac{2}{3}(25-x^2)^{3/2}+C\right]^{5}_{-5} \dots$

SteveOK…What’s going on here? Well, keep reading!

# Integrating Root Functions

Well. $\displaystyle\int_{-5}^5(x-\sqrt{25-x^2} \, dx=\int_{-5}^5 x \, dx-\int_{-5}^5\sqrt{25-x^2}\, dx$. For the 2^{nd} integral, we can’t integrate it as if it were $\sqrt{x}$. In fact, if we try to “integrate” $\displaystyle\sqrt{25-x^2}$ this way, we get:

\begin{align*} \displaystyle \frac{2}{3}\left({25-x^2}\right)^{\frac{3}{2}} \end{align*}

However,

\begin{align*} \frac{d}{dx}\left[ \frac{2}{3}\left({25-x^2}\right)^{\frac{3}{2}} \right] = \sqrt{25-x^2}(-2x) \ne\sqrt{25-x^2}\end{align*}

Oops. Looks like the obnoxious *Chain Rule* just destroyed our whole scheme.

# Troubleshooting: Our Answer

We have just seen how misusing *Power Rule* can create abnormal antiderivatives. For the 2^{nd} integral, what we need to do is to use a *back substitution* with $x=5\sin(u)$. In which case, $dx=5\cos(u)du$, and the limits of the integral now run from $\displaystyle -\frac{\pi}{2}$ to $\displaystyle \frac{\pi}{2}$. Putting everything together, we get:

\begin{align*} \int_{-5}^5 \sqrt{25-x^2}\, dx =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{25-\left(5\sin(u)\right)^2} \ 5\cos(u) \, du \end{align*}

Simplifying further, we get:

\begin{align*} \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}5\cos(u) \, 5\cos(u) \, du=25\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2(u) \, du \end{align*}

By using the *half-angle formula* for cosine $\displaystyle \left(\text{i.e., } \cos^2(u) =\frac{1 + \cos(2u)}{2} \right)$, we can continue to work on the antiderivative:

\begin{align*} 25\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2(u) \, du & = 25\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos(2u)}{2} \, du \\ & = 25\left [\frac{1}{2}x+ \frac{\sin(2u)}{4}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\ & = 25\left(\frac{\pi}{2}\right) \end{align*}

which means that our original integral, $\large \displaystyle \int_{-5}^5 x – \sqrt{25-x^2} \, dx$, would be $\displaystyle -\frac{25\pi}{2}$, as Steve has previously inferred using a geometric argument.

And if there is a moral to this story, it would be that there would be no paradox at all — as long as we don’t go for an anarchic free-for-all! 😉

Calculus Tutor NYC

May 28, 2015 @ 1:14 am

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so I came too take a look. I’m definitely enjoying the information. I’m book-marking andd will be tweeting this to my followers!

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Math Vault

May 29, 2015 @ 10:48 am

Hey many thanks! More mathematical madness has yet to come!

Chow Kim Wan

July 20, 2015 @ 8:43 pm

Hi there! I am a prospective math major and I blog about musical theatre/Disney primarily (I know that probably does not interest you), BUT I have a feature on mathematics every two Mondays or so. You might wanna check that out on my blog!

BTW this is a very common mistake I have seen among my peers. Other errors include misuse of integration by parts and not plucking in the derivative of the substitution when doing integration by substitution (they merely replace dx with du without doing anything else). When doing definite integral evaluations, they forget to change limits. Yeah, these are a few integration errors I can think of.

Math Vault

July 20, 2015 @ 9:13 pm

Wow. That’s an interesting background. We have been contemplating on writing the first textbook on what we called “mental calculus”, but it takes a bit of time and effort.

Quora

January 30, 2016 @ 6:46 pm

What is wrong with this integration technique?First, a few points: 1. [math](x^3)^2(x^2+1) = x^6 (x^2+1) = x^8 + x^6,[/math] not [math]x^{12} + x^6[/math]. 2. [math]{(x^{12} + x^6)}^{\frac{1}{2}} \ne x^{12\frac{1}{2}} + x^{6\frac{1}{2}} [/math](exponents can be distributed over products, but n…

Math Vault

January 30, 2016 @ 7:38 pm

Cool!