I only want to write this down to have a place to look it up:
If we want to integrate \(\displaystyle{ \int_{a}^{b}{f'\left(x \right) g\left( x\right)} }\) , we can use: \[ \begin{aligned} (f \cdot g)'\left( x \right) & = f'\left(x \right)g\left(x \right) + f\left(x \right)g'\left(x \right) \\ \Leftrightarrow f\left(x \right)g\left(x \right) \big | _ {a=x}^b & = \int _ {a}^{b}{ f'\left(x \right)g\left(x \right) + f\left(x \right)g' \left(x \right) } \, dx \\ \Leftrightarrow \int _ {a}^{b}{ f'\left(x \right)g\left(x \right) } \, dx & = f\left(x \right)g\left(x \right) | _ {a=x}^b - \int _ {a}^{b}{ f\left(x \right)g' \left(x \right) } \, dx \end{aligned} \]
And if we want to substitute a function in an integral:
\[{\int_{g\left(a \right)}^{g\left(b \right)}{ f\left( x \right) } \, dx = F \left ( g \left ( b \right ) \right ) - F \left ( g \left ( a \right ) \right ) = \left (F \circ g \right ) \left ( b \right ) - \left (F \circ g \right ) \left ( a \right ) = } \\=\int_{a}^{b}{ \left (F \circ g \right ) ' \left ( x \right ) } \, dx = \int_{a}^{b}{ f \left ( g \left ( x \right ) \right ) } \, g' \left ( x \right ) \, dx \]
(with \(F \left ( x \right ) = \int^{x}{ f \left ( x \right )} \, dx\)) or using a trick:
\[\int_{a}^{b}{f \left ( g \left ( x \right ) \right ) } \, dx = F \left ( b \right ) - F \left ( a \right ) = \left ( F \circ g \circ g^{-1} \right ) \left ( b \right ) - \left ( F \circ g \circ g^{-1} \right ) \left ( a \right )=\\=\left ( F \circ g^{-1} \right ) \left( g \left ( b \right ) \right ) - \left ( F \circ g^{-1} \right ) \left( g \left ( a \right ) \right ) = \int_{g \left ( a \right )}^{g \left ( b \right )}{ \left ( F \circ g^{-1} \right ) ' \left ( x \right ) } \, dx = \\=\int_{g \left ( a \right )}^{g \left ( b \right )}{ F' \left ( g^{-1} \left ( x \right ) \right ) \, {g^{-1}} ' \left ( x \right ) } \, dx = \int_{g \left ( a \right )}^{g \left ( b \right )}{ f \left ( g \left ( g^{-1} \left ( x \right ) \right ) \right ) \, {g^{-1}}' \left ( x \right ) } \, dx = \\ = \int_{g \left ( a \right )}^{g \left ( b \right )} { f \left ( x \right ) \, \frac{1}{g' \left ( g^{-1} \left ( x \right ) \right ) } } \, dx \]
(with \(F \left ( x \right ) = \int^{x}{ f \left ( g \left ( x \right ) \right )} \, dx\))