A rarely seen form of substitution of definitite integral theorem: $\int_\alpha^\beta f(u(t))dt=\int_{u(\alpha)}^{u(\beta)}f(x)\cdot(u^{-1})'(x)dx$ Below is a theorem about substitution of definite integral that I found today. However, I had never seen this form in analysis books. How to understand its meaning and usage? Is it really made used in practice? The characters for $u^{-1}$, $f(u(t))$ is messy to me. (Though I know the classic form of such theorem.)

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
  $$\int_\alpha^\beta f(u(t))dt=\int_{u(\alpha)}^{u(\beta)}f(x)\cdot(u^{-1})'(x)dx$$

Edit:
We are looking for explicitly examples where this method is usefully for computing integrals.
 A: This can be used to evaluate $\int (\tan x)^{c}dx$ for arbitrary integer $c$.
Let $0<a<b<\frac{\pi}2$.
Then, when $f(x)=x^c$, $u(x)=\tan x$, then
$$\int^b_a (\tan x)^cdx = \int^{\tan b}_{\tan a}\frac{x^{c}}{1+x^2}dx$$ which has closed form.
Another example is $f(x)=x^n$, $u(x)=W_0(x)$.
$$\int^b_a W_0(x)^n dx = \int^{W(b)}_{W(a)}x^n\cdot(e^x+xe^x)dx$$which has closed form as well.
Another example is $f(x)$ being arbitrary rational function, and $u(x)=ax^2+bx+c$.
Then, with appropriate assumptions on $c,d$,
$$2\int^{u(c)}_{u(d)}\frac{f(x)}{\sqrt{b^2-4a(c-x)}}dx=\int^d_c f(ax^2+bx+c)dx$$ where the right hand side has closed form.
You can do something similar for a cubic or quartic polynomial, and design a crazy integral to outsmart your friends by writing out a closed form without much effort:) 
Some more examples:
$$\int^b_a \sin(W_0(x))dx=\int^{W(b)}_{W(a)}\sin x(e^x+xe^x)dx=\left[\frac{e^x((x+1)\sin x-x\cos x)}2\right]^{W(b)}_{W(a)}$$
$$\int^b_a \cos(W_0(x))dx=\int^{W(b)}_{W(a)}\cos x(e^x+xe^x)dx=\left[\frac{e^x(x\sin x+(x+1)\cos x)}2\right]^{W(b)}_{W(a)}$$
$$\int^b_a \sinh(W_0(x^{1/n}))dx=\int^{u(b)}_{u(a)}\sinh x (nx^{n-1}e^{nx}+nx^ne^{nx})dx=\text{has closed form}$$
$$\int^b_a \sin\ln( x)dx=\int^{\ln b}_{\ln a}\sin (x )e^xdx=\left[\frac{e^x(\sin x-\cos x)}2\right]^{\ln b}_{\ln a}$$
In case you are still not satisfied, you can also evaluate 
$$\int^b_a\sin(p_1W(x))\sin(p_2W(x))\cdots\sin(p_AW(x))\cdot\cos(q_1W(x))\cos(q_2W(x))\cdots\cos(q_BW(x))dx$$
A: For an example where this formula can be useful, define $f:[0,1]\rightarrow\mathbb{R}, t\mapsto t$.
Then:
$$\int_0^{\pi/4}\tan(t)dt=\int_0^{\pi/4}f(\tan(t))dt=\int_{\tan(0)}^{\tan(\pi/4)}f(t)\frac{d}{dt}(\arctan(t))dt \\ =\int_{0}^{1}\frac{t}{1+t^2}dt=\frac{1}{2} {\log(2)}$$
