Substitution
The motivating example for this section will be the integral shown below.
\int \sec^2(x)e^{1 + \tan{x}} \, dx
The techniques shown in the last section don’t work here. Linearity does not allow for separation of products, and we don’t have an identity which gets us a solution. In order to solve these types of integrals, we’ll need a new rule.
The integral can be rewrote if one makes the following substitution.
u = 1 + \tan^2x
Then computing the differential
du = \sec^2x \,dx
We can put these terms into the integral and obtain
\int e^{u} \, du
Which clearly evaluates to
\int e^{u} \, du = e^u + C
And then undoing the substitution
\int \sec^2(x)e^{1 + \tan{x}} \, dx = e^{1+\tan x} + C
Substitution can often be thought of as undoing chain rule, by substituting the interior function.
The substitution rule gives
\int f(g(x))g^\prime(x)\, dx = \int f(u)\, du, \text{where } u = g(x).
The biggest problem is identitfying the function to substitute. There is no general way of identifying which function in the integrand needs to be substituted.
Examples
Evaluate the following integral
\int {{\left( {8x - 12} \right){{\left( {4{x^2} - 12x} \right)}^4}\,dx}}
Using the substitution
u = 4x^2 - 12x
We get \int {{\left( {8x - 12} \right){{\left( {4{x^2} - 12x} \right)}^4}\,dx}} = \frac{1}{5}(4x^2 - 12x)^5 + C
Evaluate the following integral
\int{{3{t^{ - 4}}{{\left( {2 + 4{t^{ - 3}}} \right)}^{ - 7}}\,dt}}
Using the substitution
u = 2 - 4t^{-3}
We get \int{{3{t^{ - 4}}{{\left( {2 + 4{t^{ - 3}}} \right)}^{ - 7}}\,dt}} = \frac{1}{24}(2+4t^{-3})^{-6} + C
Evaluate the following integral
\int \frac{\csc x \cot c}{2- \csc x} \, dx
First notice how \csc x \cot x is the derivative of -\csc x. This leads us to try some form of \csc x for u. I use
u = 2 - \csc x
Leading too \int \frac{\csc x \cot c}{2- \csc x} \, dx = - \int \frac{1}{u} \, du
Which is then
\int \frac{\csc x \cot c}{2- \csc x} \, dx = \ln |u| + C = \ln |2 - \csc x| + C
Evaluate the following integral
\int{{10\sin \left( {2x} \right)\cos \left( {2x} \right)\sqrt {{{\cos }^2}\left( {2x} \right) - 5} \,dx}}
While the integral looks incredibly complex, if we substitute the term inside the radical (the usual choice) we can start to break it down.
u = \cos^2(2x) - 5
Thus \int{{10\sin \left( {2x} \right)\cos \left( {2x} \right)\sqrt {{{\cos }^2}\left( {2x} \right) - 5} \,dx}} = -\frac{5}{2} \int \sqrt{u} \, du
Which leads too
\int{{10\sin \left( {2x} \right)\cos \left( {2x} \right)\sqrt {{{\cos }^2}\left( {2x} \right) - 5} \,dx}} = -\frac{5}{3}(\cos^2(2x) - 5)^\frac{3}{2} + C
Evaluate the following integral
\int \frac{6}{7 + y^2} \,dy
This does not follow our usual f(g(x))g^\prime(x) example. For this case, we’ll try to convert into an identity we know. It is clear we can’t have y^2 be a term in u, as this would only make the integral more complicated (as thers no y to be consumed in the du). The integrand sorta resembles the derivative of \arctan y, so we can try to rewrite it
\int \frac{6}{7 + y^2} \,dy = \frac{6}{7} \int \frac{1}{1 + \frac{y^2}{7}} \, dy
Notice that if we make the substitution u = \frac{y}{\sqrt{7}}
We get
\int \frac{6}{7 + y^2} \,dy = \frac{6\sqrt{7}}{7} \int \frac{1}{1 + u^2} \, du
Which gives us
\int \frac{6}{7 + y^2} \,dy = \frac{6\sqrt{7}}{7} \arctan \left(\frac{y}{\sqrt{7}}\right) + C
Evaluate the previous integral in it’s general form
\int \frac{1}{a^2 + x^2} \,dx
Like last time, we’ll factor out a^2 from the bottom and then use a substitution.
\int \frac{1}{a^2 + x^2} \,dx = \frac{1}{a^2} \int \frac{1}{1 + (x^2/a^2)} \, dy
Makeing the substitution u = \frac{x}{a}
We get
\int \frac{1}{a^2 + x^2} \,dx = \frac{1}{a} \int \frac{1}{1 + u^2} \, dx
Which gives us
\int \frac{1}{a^2 + x^2} \,dx = \frac{1}{a} \arctan \left(\frac{x}{a}\right) + C
Evaluate the following integral
\int{{20{{e}^{2 - 8w}}\sqrt {1 + {{e}}^{2 - 8w}}} \, + 7{w^3} - 6\,\,\sqrt[3]{w}\,dw}
Using linearity on the integral allows us to do different substitutions. Seeing as the last two terms can be integrated with the inverse power rule, we only need to a substitution on the first term. Luckily this is an easy case.
u = 1 + e^{2-8w}
Gives
\int{{20{{e}^{2 - 8w}}\sqrt {1 + {{e}}^{2 - 8w}}} \,dw} = -\frac{5}{2}\int \sqrt{u}\, du
Which is
\int{{20{{e}^{2 - 8w}}\sqrt {1 + {{e}}^{2 - 8w}}} \,dw} = -\frac{5}{3}(1 + e^{2-8w})^\frac{3}{2} + C
Then solving the other terms gives us our solution.
= -\frac{5}{3}(1 + e^{2-8w})^\frac{3}{2} + \frac{7}{4}w^4 - \frac{9}{2}w^\frac{4}{3} + C
Evaluate the following integral
\int{{\frac{{{x^7}}}{{\sqrt {{x^4} + 1} }}\,dx}}
Using the substitution
u = x^4 + 1
We get
\int{{\frac{{{x^7}}}{{\sqrt {{x^4} + 1} }}\,dx}} = \frac{1}{4}\int \frac{u - 1}{\sqrt{u}}
Which we can then apply linearity and inverse power rule to, giving us
\int{{\frac{{{x^7}}}{{\sqrt {{x^4} + 1} }}\,dx}} = \frac{1}{4}\left( \frac{2}{3}(x^4 + 1)^\frac{3}{2} + 2(x^4 + 1)^\frac{1}{2}\right) + C