I know that this is a really elementary question, but I have come to realize that I am thinking single variable and multivariable calculus as two different topics in the sense that 99% of the time I have "just used" the u-substitution method without really thinking how it reflects to the more general Change of variables theorem. In particular, I am not really sure how to use the said theorem precisely when evaluating even a simple integral like $I = \int \tan(x)dx$.

The $u$-substitution method yields $u = \cos(x)$ we get $\int \tan(x)dx = \int \frac{\sin(x)}{\cos(x)}dx = \int -\frac{du}{u} = -\mathrm{ln}(|u|) + C$.

But the change of variables theorem requires using a specific mapping $\varphi$. I suppose that in this case $x = \varphi(u) = \cos^{-1}(u)$. Wouldn't we then have something like $f(x) = \tan(x) = \tan(\varphi(u)) = \tan(\cos^{-1}(x))$? In particular, I am curiour of knowing how the $\int_{\varphi(M)} f(v)dv = \int_Mf(\varphi(u))\left|\mathrm{det}(D\varphi)(u)\right|du$ looks in this situation. Surely $\left|(D\varphi)(u)\right| = \left|\frac{-1}{\sqrt{1 - u^2}}\right|$, but beyond this I am not sure how to navigate the computations (w/o just resorting to the original substitution).



2 Answers 2


This is going to be a more convoluted way of solving this integral, but bare with me.

Note first that the theorem is talking about definite integrals, and you are trying to come up with an answer to an indefinite integral.

Then note that the theorem is called $\textit{change of variables}$, not $\textit{change of functions}$ or $\textit{change of integrands}$. It turns out that $u$-substitution works, and follows from the same premises, but it's application is slightly different.

Here's how to solve this integral by applying a change of variables, namely, changing $x$.

Set $x = arctan(u)$. Differentiating gets you $$\frac{dx}{du}=\frac{1}{1+u^2}\implies dx = \frac{1}{1+u^2}du.$$

Substituting, the integral is now $$\int{\frac{tan(arctan(u))}{1+u^2}}du.$$ The numerator is simply $u$, so you multiply by $1$ in the form $\frac{2}{2}$ and pull the 2 in the numerator into the integral using the property of homogeneity.

You end up with $$\frac{1}{2}\int{\frac{2u}{u^2+1}}du = \frac{1}{2}log(u^2+1) + C.$$

Note that your initial substitution $x = arctan(u)$ implies $u=tan(x)$.

Substituting, you get $$\frac{1}{2}log(u^2+1) = \frac{1}{2}log(tan^2(x)+1) = \frac{1}{2}log\left(\frac{sin^2(x)}{cos^2(x)}+1\right).$$

Now express the $1$ as $\frac{cos^2(x)}{cos^2(x)}$ to obtain $$\frac{1}{2}log\left(\frac{cos^2(x) + sin^2(x)}{cos^2(x)}\right) = \frac{1}{2}log\left(\frac{1}{cos^2(x)}\right) = \frac{1}{2}log(cos^{-2}(x)) = -log(cos(x)),$$ the last step utilizing the exponent property of the logarithm.

So finally, you have $$\int{tan(x)}dx = -log(cos(x))+C.$$

Now to relate this back to your question:

we substituted $x = \phi(u) = arctan(u)$, and differentiation gave us $\frac{dx}{du}=\phi'(u)=\frac{1}{1+u^2}$.

So you can see $$ \int{f(\phi(u))\phi'(u)}du = \int{tan(arctan(u))\cdot\frac{1}{1+u^2}}du = \int{\frac{u}{u^2+1}}du.$$

Phew. So this was a solution we got by $\textit{changing the variable x}$. It might be instructive to think about how this differs from $u$-substitution, why $u$-substitution works (hint: does $x = arctan(u)\implies u=tan(x)$ work in the other direction?), and what it's doing.

As far as the Jacobian goes, in the one-variable scenario that's just what you're doing by accounting for the $dx$ when preforming the substitution. It turns out that this area of Analysis has some very inspiring connections to Linear Algebra, so you might be able to relate your knowledge of that subject to this theorem. There are countless resources that will give you better "intuition" on what's going on here than anything I can write in this post, for example this YouTube video.


It is not so hard. You set $u=cosx$ then $\dfrac{du}{dx}=-sinx$ and hence

$-du=sinxdx$ so the integral becomes $\int\dfrac{-du}{u}$.

  • $\begingroup$ But shouldn't we have absolute values around the derivative of $u$ w.r.t. $x$, since we are taking the absolute value of the determinant of the Jacobian? $\endgroup$ Commented Sep 7, 2022 at 7:10
  • $\begingroup$ When we make change of variables we limit ourselves on an interval such that the transformation is $1-1$. In our example the interval is $[0,\pi]$ so we have $u:[0,\pi]\\,\to\,[-1,1]$. The absolute value is only necessary in the solution of the integral! $\endgroup$
    – user1054388
    Commented Sep 7, 2022 at 7:17

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