HỌC LẠI LOGARIT TỪ ĐẦU ^^ – TOÁN 12 – THẦY NGUYỄN CÔNG CHÍNH
HỌC LẠI LOGARIT TỪ ĐẦU ^^ – TOÁN 12 – THẦY NGUYỄN CÔNG CHÍNH

Trig Integrals, Integral Calculus,, Calculus 2, Calculus II, McGill MATH 122, McGill MATH 139, McGill MATH 140, McGill MATH 141, McGill MATH 150, Concordia MATH 203, Math Tutoring, Montreal

(For an integration method which can potentially substitute integration by parts and the substitution method itself, see The Overshooting Method.)

Yes. It can be sometimes amusing to see calculus students mis-integrating 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: $\displaystyle \int^5_{-5}\left(x-\sqrt{25-x^2}\right)\,dx=\left[\frac{x^2}{2}-\frac{2}{3}(25-x^2)^{3/2}+C\right]^{5}_{-5} \dots$

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

Table of Contents

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 2nd 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:

\[ \frac{2}{3}\left({25-x^2}\right)^{\frac{3}{2}} \]

However,

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

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

Integration Failure, Calculus 2, Trig Integral, McGill, Concordia, Math Tutor, Tutoring, Montreal, MATH 141, MATH 203

Troubleshooting: Our Answer

So is there a paradox? Not really. It’s merely an illustration of how the misuse of Power Rule can create abnormal antiderivatives. For the 2nd 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 (i.e., $\cos^2 u =\left(1 + \cos(2u)\right)/2\,$), 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! 😉

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