# What's the practical situation to use the integration by substitution like this?

I saw a weird form of integration by substitution for definite integrals in an analysis textbook. This theorem is not in the exercise, but rather in the normal paragraph. The form of this one seems so strange to me. I want to ask what is the situtation we need to use this kind of substitution. I had a moment guess that this may be used involve triangometric substitution, but the pattern of such situtation still seems quite different from this one.

Let $J=[\alpha,\beta],~u:J\to\Bbb R$ be a $C^1$ function and $u'(x)\neq 0$ for all $x\in J$, $I$ be an interval and $u(J)\subseteq I$, $f:I\to\Bbb R$ be continuous. Then $$\int_\alpha^\beta f(u(t))dt=\int_{u(\alpha)}^{u(\beta)}f(x)\cdot(u^{-1})'(x)dx$$

• What's so weird about it? It's just regular substitution, but $u$ is on the other side. – cronos2 Nov 24 '17 at 14:16
• What's important here is to learn to recognize something you already know hidden in something that looks weird or strange, as is shown in the answer of @StellaBiderman. – Lee Mosher Nov 24 '17 at 14:58
• Along the lines of @LeeMosher comment, see the text I added at the end about where my example applications came from. – Stella Biderman Nov 24 '17 at 15:38

This is regular u-sub, just written a little differently. Regular u-substitution says that $$\int f(x)dx=\int f(u(t))u’(t)dt$$ Now substitute $t=u^{-1}(x)$ to give us $$\int f(u(t))dt=\int f(u(u^{-1}((x)))u’(u^{-1}(x))dx=\int f(x)(u^{-1})’(x)dx$$ where the last equality follows from the inverse function theorem (check for yourself that the bounds end up correct).
$$\int(\sin^{-1}(x))^2dx\quad\int \log(\sqrt{x})dx\quad \text{and }\int 3(\log(x))^4dx$$
Where did I get these expression from? They all came from standard applications of “normal u-substitution” except modified to use inverse functions so that the term $u^{-1}’$ turns into a more familiar function.
• Why is $\int f(u(t))dt=\int f(u(u^{-1}((x)))u’(u^{-1}(x))dx$? (I sometimes have trouble playing with the $dx,du,\cdots$ symbols.) – Eric Nov 24 '17 at 14:51
• @Eric Think of $f(u(\cdot))$ as a single function (maybe call it $g(t)$). Then the first equality comes from taking $\int g(t)dt$ and applying the normal u-substitution rule with $t=u^{-1}(x)$. Notice that this is the same equation as $u(t)=x$, just inverted. – Stella Biderman Nov 24 '17 at 15:00