Step By Step Calculus » 15.2 - Change of Variables

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Synopsis
It is often useful to break down the process of finding an antiderivative into parts using the chain rule. Suppose FF is an antiderivative for f.f. Then,

\displaystyle \newcommand{\stackunder}[2]{\mathrel{\mathop{#2}\limits_{#1}}}{ \textbf{(Change of Variable)}\qquad\int \stackunder{f(t)}{\underbrace{f(g(x))}}\stackunder{dt}{\underbrace{g^{\prime }(x)dx}}=F(g(x))+c=\left. \int f(t)dt\right| _{t=g(x)} }\displaystyle \newcommand{\stackunder}[2]{\mathrel{\mathop{#2}\limits_{#1}}}{ \textbf{(Change of Variable)}\qquad\int \stackunder{f(t)}{\underbrace{f(g(x))}}\stackunder{dt}{\underbrace{g^{\prime }(x)dx}}=F(g(x))+c=\left. \int f(t)dt\right| _{t=g(x)} }

Here we are presented with the integral on the left. Recognize f,gf,g on t,t, then integrate just ff and substitute t=g(x)t=g(x) back in. This method is known as a change of variable (or sometimes the substitution method). We can use one of the following two procedures to solve integrals that requires change of variables. The first one is the MESS procedure for substitution. This procedure is particularly useful for the problems where finding g'(x)g'(x) in the integrand is easy. It has the following steps:
(Manipulate) Manipulate the integral to find g(x)g(x) of f(g(x))f(g(x)) and/or g'(x)dxg'(x)dx.
(Eliminate) Express the integral in terms of tt by substituting with t=g(x)t=g(x) and dt=g'(x)dxdt=g'(x)dx.
(Solve) Solve \displaystyle \int f(t) dt \displaystyle \int f(t) dt .
(Substitute) Substitute t=g(x)t=g(x) back into answer.
The second one is the ADRESS procedure for substitution. This procedure is particularly useful for the problems where finding g'(x)g'(x) in the integrand is hard. It has the following steps:
(Assign) Choose g(x)g(x).
(Differentiate) Compute g'(x)g'(x).
(Ratio) Multiply the integrand with ratio \dfrac{g'(x)}{g'(x)}\dfrac{g'(x)}{g'(x)}.
(Eliminate) Express the integral in terms of tt by substituting with t=g(x)t=g(x) and dt=g'(x)dxdt=g'(x)dx. If it is not possible to express the integral in terms of tt, then we have made a bad choice for g(x)g(x). In that case, try a different expression for g(x)g(x).
(Solve) Solve \displaystyle \int f(t) dt \displaystyle \int f(t) dt .
(Substitute) Substitute t=g(x)t=g(x) back into answer.