# Indefinite integrals and changing variables

Why is it legal to change the variable of an indefinite integral?

Consider $$\int \dfrac{dx}{\cos x}$$

If one were to say, $\text{Let } u=\cos x$, do we not now technically have $$\int_{u=\cos-\infty}^{u=\cos\infty}u^{-1}\dfrac{d\arccos u}{du} \cdot du$$ Which is clearly madness even before we consider that whatever $\cos \infty$ is, $\cos -\infty$ is too.

So my question is simply, what have I missed; why is it allowed?

• If you consider $u=\cos x$ then $x=\arccos u.$ So $dx=\frac{-1}{\sqrt{1-u^2}}du.$ Where is $\arccos u$ in the integral? On the other hand, why you try to convert and indefinite integral into a definite integral? – mfl May 13 '14 at 19:21
• Woops. That was a dumb error. Well an indefinite is no different to 'definite over infinite range in each direction' right? – OJFord May 13 '14 at 19:28
• They are completely different things. An indefinite integral is a function (if we assume some normalization on the constant of integration) and a definite integral is a number (if it exits). – mfl May 13 '14 at 19:31
• But am I wrong in saying $\int_{-\infty}^{\infty}y(x)dx := \int y(x)dx$? – OJFord May 13 '14 at 19:39
• I'm not "trying to turn it into a definite integral" - I'm just asking why it is allowed to change the variable. If it was definite, the values of those bounds would also change. – OJFord May 13 '14 at 19:41

You don't even hint at the nature of the mysterious process that led you to the conclusion that the bounds of integration would be $\pm\infty$. Let's try it with some actual bounds: $$\int_0^{\pi/4}\frac{dx}{\cos x} = \int_1^{\sqrt{2}/2} \frac{d\arccos u}{u} = \int_1^{\sqrt{2}/2} \frac{-du}{u\sqrt{1-u^2}}$$ No "madness" is involved.

If you want to think about $\displaystyle\int_{-\infty}^\infty \frac{dx}{\cos x}$, you'll meet some unpleasant complications. $$\int_{-\infty}^\infty \frac{dx}{\cos x} = \sum_{n=-\infty}^\infty \int_{2\pi n}^{2\pi(n+1)} \frac{dx}{\cos x} = \sum_{n=-\infty}^\infty \int_0^{2\pi} \frac{dx}{\cos x}.$$ That last sum would be $0$ if the value of the integral is $0$ and either $\pm\infty$ if it is any other number. So let's think about $$\int_0^{2\pi} \frac{dx}{\cos x} = \left(\int_{-\pi/2}^{\pi/2} + \int_{\pi/2}^{3\pi/2} \right) \frac{dx}{\cos x} = \infty + (-\infty).$$ Perhaps one could say that a sort of Cauchy principal value of this integral is $0$. At each of the two locations $a$ of vertical asyptotes, one would have $$\lim_{\varepsilon\downarrow0}\left(\int_\cdots ^{a-\varepsilon} + \int_{a+\varepsilon}^ \cdots\right).$$ Then one would get $0$.

But one should avoid substitutions that are not one-to-one.

As to why substitutions are valid, the short answer, which you'll find in every textbook is: the chain rule.

• When I was first taught to integrate, I was told $\int = \int_{-\infty}^{\infty}$. I never really thought too much about it again until now. – OJFord May 13 '14 at 20:41
• I never said there was anything mad about the definite integral, just misunderstood the indefinite. – OJFord May 13 '14 at 20:42
• Further, I clearly understood that the chain rule allows substitutions of definite integrals, since I applied it in my OP. There's really no need for such condescension.. – OJFord May 13 '14 at 20:44
• What you say you were told I've never seen before. – Michael Hardy May 13 '14 at 20:45
• Okay, sorry I misunderstood.. It was a while ago, I think we were taught definite first (as the application to area under curve was emphasised far more than proof), so from there I guess we were told indefinite "is like definite, but from $-\infty$ to $\infty$" or such. I don't know. – OJFord May 13 '14 at 20:52