Integration by Substitution

Theorem
Let $$\phi$$ be a real function which has a derivative on the closed interval $$\left[{a \,. \, . \, b}\right]$$.

Let $$I$$ be an open interval which contains the image of $$\left[{a \,. \, . \, b}\right]$$ under $$\phi$$.

Let $$f$$ be a real function which is continuous on $$I$$.

Then $$\int_{\phi \left({a}\right)}^{\phi \left({b}\right)} f \left({t}\right) dt = \int_a^b f \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right) du$$.

Proof
Let $$F \left({x}\right) = \int_{\phi \left({a}\right)}^x f \left({t}\right) dt$$.

From Derivative of a Composite Function and the first part of the Fundamental Theorem of Calculus:

$$\frac d {du} F \left({\phi \left({u}\right)}\right) = F^{\prime} \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right) = f \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right)$$.

Thus from the second part of the Fundamental Theorem of Calculus:

$$\int_a^b f \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right) du = \left[{F \left({\phi \left({u}\right)}\right)}\right]_a^b$$

which is what we wanted to prove.