Mathematics-Online course: Prepcourse Mathematics-Analysis-Integral Calculus-Substitution of Variables

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	Mathematics-Online course: Prepcourse Mathematics - Analysis - Integral Calculus
	Substitution of Variables

From the chain rule

$\displaystyle \frac{d}{dx} F(g(x))=f(g(x)) g'(x),\quad f=F',$

follows by integration

$\displaystyle \int f(g(x)) g'(x) dx = F(g(x)) +c.$

According to definite integrals holds

$\displaystyle \int_a^b f(g(x)) g'(x) dx = F(g(b))-F(g(a)) = \int_{g(a)}^{g(b)}f(y) dy.$

By means of differentials one write this formula in the form

$\displaystyle \int_a^b f(y) \frac{dy}{dx}dx = \int_{g(a)}^{g(b)} f(y)dy.$

To calculate $\int f(x)dx$ one substitutes

$\displaystyle x=g(t) \qquad {\mbox{und}} \qquad dx=g'(t)dt,$

with an appropriate invertible function

, and calculates

$\displaystyle \int f(g(t))g'(t)dt = H(t) +c.$

By inversion of the function

, i.e.

, one obtains

$\displaystyle \int f(x)dx = H(h(x))+c.$

For example

$\displaystyle F(x)= \int \frac{e^{3x}}{e^{2x}-1}dx$

can be determined by means of substution $x=\ln t$ and $dx=\frac{1}{t}dt$ :

$\displaystyle F(\ln (t))$	$\displaystyle =$	$\displaystyle \int \frac{t^2}{t^2-1}dt = \int \left( 1+\frac{1}{t^2-1} \right) dt$
	$\displaystyle =$	$\displaystyle \int dt + \int \frac{1}{t^2-1}dt$
	$\displaystyle =$	$\displaystyle t + \frac{1}{2} \ln \left\vert \frac{t-1}{t+1} \right\vert +c$

By resubstitution of

results

$\displaystyle F(x)=e^x+\frac{1}{2} \left\vert \frac{e^x-1}{e^x+1} \right\vert+c.$

automatically generated 9/18/2007