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 for integrals is often in the form
\int f(g(x))g^\prime(x)\, dx = \int f(u)\, du, \text{where } u = g(x).
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 clearly the derivative of -\csc x. This leads us to try some form of \csc x for u. I use
u = \csc x
Leading too \int \frac{\csc x \cot c}{2- \csc x} \, dx = - \int \frac{1}{2 - u} \, du
Which is clearly
\int \frac{\csc x \cot c}{2- \csc x} \, dx = \ln |2 - 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