BlackHC’s Adventures in the Dev World

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:

$\displaystyle { (f \cdot g)'\left( x \right) & = f'\left(x \right)g\left(x \right) +  f\left(x \right)g'\left(x \right) } \\
 \displaystyle { \Leftrightarrow f\left(x \right)g\left(x \right) \big |_{a=x}^b } & = \displaystyle { \int_{a}^{b}{ f'\left(x \right)g\left(x \right) +  f\left(x \right)g' \left(x \right) } \, dx  } \\
\displaystyle { \Leftrightarrow \int_{a}^{b}{ f'\left(x \right)g\left(x \right) }\, dx } & \displaystyle { = f\left(x \right)g\left(x \right) |_{a=x}^b - \int_{a}^{b}{ f\left(x \right)g' \left(x \right) } \, dx }$

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 $F \left ( x \right ) = \int^{x}{ f \left ( g \left ( x \right ) \right )} \, dx$)