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I'm trying to calculate $$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\sin t \cos^3 t\,dt$$ using integration by substitution $$\int_{\varphi([a;b])} f(x)dx=\int_{[a;b]} f\left(\varphi(t)\right)|\varphi'(t)|dt$$

First Method
Let $\varphi(t)=\cos t$ which is continuously differentiable and $\displaystyle\varphi\left(\left[\frac{-\pi}{2};\frac{\pi}{2}\right]\right)=[0;1]$ so $$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\sin t \cos^3 t\,dt=\int_0^1x^3dx=\frac{1}{4}$$

Second Method
Let $\varphi(t)=\cos t$ which is continuously differentiable and $\displaystyle\varphi\left(\left[\frac{-\pi}{2};\frac{\pi}{2}\right]\right)=[0;1]$ so $$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\sin t \cos^3 t\,dt=\int_{\frac{-\pi}{2}}^{0}\sin t \cos^3 t\,dt+\int_0^{\frac{\pi}{2}}\sin t \cos^3 t\,dt=\\ -\int_{\frac{-\pi}{2}}^0 f\left(\varphi(t)\right)|\varphi'(t)|dt+\int_0^{\frac{\pi}{2}} f\left(\varphi(t)\right)|\varphi'(t)|dt=-\int_0^1x^3dx+\int_0^1 x^3dx=0$$

The answer is obviously $0$ as the integrand is an odd function. As mentioned in another question, $\varphi$ isn't required to be monotone or injective. However, most high school teachers I know and most school textbooks warn students that you should not use integration by substitution if $\varphi$ is not monotone.

Do you think I should go with their methods or use the second method (with absolute values) even though it may confuse many students as they usually do it like the first method.

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  • $\begingroup$ I think one should teach $u$-substitution as a method to find antiderivatives instead of a method to evaluate integrals. (I.e in this problem use $f'(\phi(t))\phi '(t)=(f\circ \phi)'(t)$ to find an antiderivative for $\cos ^3(t)\sin t$, and then use the fundamental theorem of calculus to find the definite integral). I cannot believe most school textbooks would warn students to "...not use integration by substitution if $\phi$ is not monotone." Anyone who says this is plain incorrect. $\endgroup$ – PVAL Aug 11 '15 at 19:01
  • $\begingroup$ @PVAL That's an example from a textbook I have right now. It's in french. I says that the $u$-substitution isn't valid if $\varphi$ isn't monotone. $\endgroup$ – Paracosmiste Aug 11 '15 at 19:08
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    $\begingroup$ I suggest you look carefully at the two formulas given in user52817's answer. The first does not require that $\phi$ be monotone, and this is the formula given on the English-language Wikipedia page you've linked to. The second does require that $\phi$ be monotone, and this is the formula given on the French-language page you've linked to. Those are two different versions of "integration by substitution", one requiring monotonicity and the other not. $\endgroup$ – Santiago Canez Aug 12 '15 at 18:49
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    $\begingroup$ @SantiagoCanez Things were mixed up. I understand the difference now that I read the proof. $\endgroup$ – Paracosmiste Aug 12 '15 at 19:16
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    $\begingroup$ I should point out why these two versions exist: The version with absolute values, and intervals of integration is appropriate for integration of a measure, while the other version without an absolute value is appropriate for the integration of a differential form. Generally we emphasize the differential form approach in beginning calculus (without using those words). $\endgroup$ – Steven Gubkin Aug 14 '15 at 13:21
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There are two formulations for definite integrals:

$$\int_{\phi(a)}^{\phi(b)} f(x)\, dx=\int_a^b f(\phi(t))\phi'(t)\, dt$$

and the one you state:

$$\int_{\phi([a,b]}f(x)\,dx=\int_{[a,b]} f(\phi(t))|\phi'(t)|\, dt$$ In the second, you do need $\phi$ to be monotone. In the first formulation, you do not need this assumption. Of course when you apply the first formulation to your integral, $\phi(a)=\phi(b)$ so you get $0$.

I think the first formulation is the one typically stated in high school and undergraduate calculus textbooks.

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  • $\begingroup$ I think $\varphi$ should be monotone in the first one not the second as my example above. $\endgroup$ – Paracosmiste Aug 11 '15 at 23:51
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    $\begingroup$ @whatever Using the second version with $\phi$ not monotone is what caused the erroneous $1/4$ in your question. If you re-do that calculation with user52817's first version, you'll get the right answer $0$. $\endgroup$ – Andreas Blass Aug 12 '15 at 0:56
  • $\begingroup$ @AndreasBlass I got a wrong answer because I didn't use the absolute value not because user52817's 2nd method is wrong. In fact his 2nd method is correct, the 1st one is wrong if $\varphi$ isn't increasing on $[a;b]$. $\endgroup$ – Paracosmiste Aug 12 '15 at 11:29
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    $\begingroup$ @whatever, no, you did not get a wrong answer in your first approach because you didn't use absolute values---you got a wrong answer since replacing the given bounds with $0$ and $1$ isn't valid since $\phi$ is not monotone on $[-\pi/2,\pi/2]$. Here are you trying to use the second method in this answer and the fact that $\phi([-\pi/2,\pi/2]) = [0,1]$. If you had used the first method in this answer your bounds would have been $\phi(-\pi/2)$ and $\phi(\pi/2)$, which are both zero and hence gives you the correct value of $0$ for the integral. $\endgroup$ – Santiago Canez Aug 12 '15 at 15:13

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