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[parent] a lecture on integration by substitution (Feature)

The Method of Substitution (or Change of Variables)

The following is a general method to find indefinite integrals that look like the result of a chain rule.

  • When to use it: We use the method of substitution for indefinite integrals which look like the result of a chain rule. In particular, try to use this method when you see a composition of two functions.
  • How to use it: In this method, we go from integrating with respect to $ x$ to integrating with respect to a new variable, $ u$, which makes the integral much easier.
    1. Find inside the integral the composition of two functions and set $ u=$ “the inner function”.
    2. We also write $ du=\frac{du}{dx}dx$.
    3. Substitute everything in the integral that depends on $ x$ in terms of $ u$.
    4. Integrate with respect to $ u$.
    5. Once we have the result of integration in terms of $ u$ ($ + C$), substitute back in terms of $ x$.

The method is best explained through examples:

Example 0.1   We want to find $ \int e^{2x} dx $. The integrand is $ e^{2x}$, which is a composition of two functions. The inner function is $ 2x$ so we set:
$\displaystyle u=2x,\quad du=2dx$
Thus,
$\displaystyle x=u/2,\quad dx=du/2$
Substitute into the integral:
$\displaystyle \int e^{2x}dx= \int e^u \frac{du}{2}=\frac{1}{2}\int e^u du = \frac{1}{2} e^u +C=\frac{1}{2}e^{2x} + C$

The following are typical examples where we use the subsitution method:

Example 0.2  
$\displaystyle \int xe^{3x^2+7} dx$
The inner function is $ u=3x^2+7$ and $ du=6x dx$. Thus $ dx=du/(6x)$. Substitute:
$\displaystyle \int xe^{3x^2+7}dx = \int \frac{x e^u }{6x}du =\int \frac{e^u}{6} du = \frac{e^u}{6} +C = \frac{e^{3x^2+7}}{6} +C.$
Example 0.3  
$\displaystyle \int \sin (3x+7) dx$
The inner function is $ u=3x+7$ and $ du=3 dx$. Therefore:
$\displaystyle \int \sin (3x+7) dx= \int \frac{\sin(u)}{3} du = -\frac{\cos(u)}{3} +C = -\frac{\cos(3x+7)}{3}+C.$
Example 0.4  
$\displaystyle \int (2x+3)\sqrt{x^2+3x+20}\ dx $
Inner $ u=x^2+3x+20$ and $ du=(2x+3)dx$. Thus:
$\displaystyle \int (2x+3)\sqrt{x^2+3x+20}\ dx = \int \sqrt{u} du= \int u^{1/2} du = \frac{2u^{3/2}}{3} + C = \frac{2(x^2+3x+20)^{3/2}}{3} +C.$

Now another integral which is a little more difficult:

Example 0.5  
$\displaystyle \int \frac{\cos (\ln x)}{x} dx$
The inner function here is $ u=\ln x$ and $ du= \frac{1}{x} dx$.
$\displaystyle \int \frac{\cos(\ln x)}{x} dx = \int \cos (u) \cdot \frac{1}{x} dx= \int \cos (u) du = \sin (u) +C= \sin(\ln x) +C.$
Example 0.6  
$\displaystyle \int \frac{3x^2+14x+1}{x^3+7x^2+x+115} dx $
This function is also a typical example of integration with substitution. Whenever there is a fraction, and the numerator looks like the derivative of the denominator, we set $ u$ to be the denominator:
$\displaystyle u=x^3+7x^2+x+115, \quad du = (3x^2 +14x +1) dx$
Thus:
$\displaystyle \int \frac{3x^2+14x+1}{x^3+7x^2+x+115} dx = \int \frac{1}{u} du = \ln u + C= \ln (x^3+7x^2+x+115) + C.$
Example 0.7  

$\displaystyle \int \frac{7}{1+3x} dx $
As in the example above, we set $ u=1+3x$, $ du=3 dx $:
$\displaystyle \int \frac{7}{1+3x} dx = \int \frac{7}{u} \frac{du}{3} = \frac{7}{3} \int \frac{1}{u} du = \frac{7}{3} \ln u + C = \frac{7}{3} \ln (1+3x) + C.$
Example 0.8  
$\displaystyle \int t^3(t^4-50)^{700} dt$
Here the inner function is $ u=t^4-50$ and $ du=4t^3 dt$. Thus
$\displaystyle \int t^3(t^4-50)^{700} dt = \int \frac{u^{700}}{4} du=\frac{1}{4} \frac{u^{701}}{701} +C= \frac{(t^4-50)^{701}}{4\cdot 701} +C.$

Some other examples (solve them!):

$\displaystyle \int e^x \sin (e^x) dx,\quad \int \frac{e^x}{e^x+1} dx, \quad \int \frac{1}{x\ln x } dx$



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See Also: a lecture on integration by parts, A lecture on trigonometric integrals and trigonometric substitution, a lecture on the partial fraction decomposition method


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Cross-references: denominator, derivative, numerator, fraction, inner, inner function, terms, integral, variable, functions, composition, chain rule, indefinite integrals
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This is version 1 of a lecture on integration by substitution, born on 2006-01-26.
Object id is 7573, canonical name is ALectureOnIntegrationBySubstitution.
Accessed 7304 times total.

Classification:
AMS MSC: 26A36 (Real functions :: Functions of one variable :: Antidifferentiation)
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