Section 12.3 Integrals and the Method of Substitution
ΒΆSubsection 12.3.1 Substitution and Antiderivatives
To apply the method of substitution, we start with an integral whose integrand is a function the independent variable (x) which appears to involve a composition (suggesting a chain rule). Define u to be the formula in the composition and compute uβ². We then substitute dx=duuβ² in the integral and attempt to rewrite the entire integrand in terms of only u. We then find antiderivatives in terms of u and express the result in terms of the orignal variable.Example 12.3.1.
Use the method of substitution to find β«e3xdx.
The integrand \(e^{3x}\) involves composition with \(u=3x\text{.}\) This is our substitution variable. Because \(u'=3\text{,}\) we have \(du = 3 dx\) so that \(\displaystyle dx = \frac{du}{3}\text{.}\) We rewrite the integral in terms of the substitution variable \(u\text{.}\) After antidifferentiation using the variable \(u\text{,}\) we back-substitute our original formula for \(u=3x\text{.}\) The work is shown below.
Example 12.3.2.
Use the method of substitution to find β«β3x+5dx.
The integrand \(\sqrt{3x+5}=(3x+5)^{1/2}\) involves composition with \(u=3x+5\text{.}\) Because \(u'=3\text{,}\) we have \(du = 3 dx\) and \(\displaystyle dx = \frac{du}{3}\text{.}\) We rewrite the integral in terms of \(u\text{,}\) find an antiderivative, and then back-substitute to find a formula in terms of \(x\text{.}\)
Example 12.3.3.
Use the method of substitution to find β«xsin(x2)dx.
The integrand \(x \sin(x^2)\) is a product with the composition involving \(u=x^2\text{.}\) We hope that the product is a result of the chain rule. Because \(u'=\frac{du}{dx} = 2x\text{,}\) we have \(\displaystyle du = 2x \, dx\) or \(\displaystyle dx = \frac{du}{2x}\text{.}\) We rewrite the integral
This problem relied on the factor \(x\) and the formula for \(u'=2x\) having \(x\) cancel so that the transformed integral involves only the substitution variable \(u\text{.}\)
Example 12.3.4.
Use the method of substitution to find β«tan(x)dx.
The integrand \(\tan(x)\) can be rewritten as a quotient, or as a product involving a negative power,
Once we have the negative power, we see the composition variable \(u=\cos(x)\text{.}\) Because \(u'=\frac{du}{dx} = -\sin(x)\text{,}\) we have \(\displaystyle du = -\sin(x) \, dx\) or \(\displaystyle dx = \frac{-du}{\sin(x)}\text{.}\) We rewrite the integral
Example 12.3.5.
Use the method of substitution to find β«xβ1βxdx.
The integrand \(x \sqrt{1-x} = x (1-x)^{1/2}\) is a product with the composition involving \(u=1-x\text{.}\) Because \(u'=\frac{du}{dx} = -1\text{,}\) we have \(\displaystyle du = -dx\) or \(\displaystyle dx = -du\text{.}\) We rewrite the integral
If we start with the substitution equation \(u=1-x\) and solve for \(x\text{,}\) we find \(x=1-u\) and can use this substitution in the integral.
As currently written in a product, the antiderivative can not be found. However, if we multiply this out we can find antiderivatives using the power rule.
Example 12.3.6.
Use the method of substitution to rewrite β«eβx2dx.
The integrand \(e^{-x^2}\) has a composition involving \(u=x^2\) and \(u'=2x\text{.}\) For \(x \gt 0\) we have the back-substitution
The method of substitution allows us to rewrite this integral.
While these integrals are equivalent for \(x \gt 0\text{,}\) the new integral is no easier to evaluate than the original. It happens that this integral does not have an elementary antiderivative formula.
Subsection 12.3.2 Substitution and Definite Integrals
When using definite integrals, the Fundamental Theorem of Calculus allows us to compute a definite integral as the change in an antiderivative. If the method of substitution is used, our antiderivative will be a function of the substitution variable u which is a function of the independent variable. Rather than rewrite the antiderivative in terms of the original variable and then compute the change of the antiderivative, we can compute the change in the antiderivative in terms of the variable u. Suppose that F(x) is an antiderivative of f(x). Now, suppose that u is a function of x so that u(a)=c and u(b)=d. If we have an integral involving composition and the chain rule, we findExample 12.3.7.
Compute β«31(2x+1)4dx.
The substitution variable is \(u=2x+1\text{.}\) When \(x=1\text{,}\) \(u=2(1)+1=3\text{,}\) and when \(x=3\text{,}\) \(u=2(3)+1=7\text{.}\) The substitution step involves \(u'=\frac{du}{dx}=2\) so that \(du = 2dx\) or \(\displaystyle dx=\frac{du}{2}\text{.}\) In order to keep track of whether the limit of integration refers to \(x\) or \(u\text{,}\) we need to clearly indicate this when both variables are involved.
Example 12.3.8.
Compute β«43xdx25βx2.
The composition may not be apparent until we think of division as multiplication by a negative power:
This suggests a substitution \(u=25-x^2\text{.}\)