Solving Integrals Through the Use of Integration by Parts

Overview

Suppose we have an integral which appears unsolvable by ordinary means, such as
∫xsin(x)dx In many cases, we can do what is called integration by parts, which is where we split the equation we are taking the integral of into two parts, with the goal of simplifying the equation such that we can reduce it down to an equation we know how to solve.

Formula

Figure 1: General formula for Integration by Parts

In essence, what this means is that we are going to attempt to split up the equation in a way such that we can eliminate one of the equations inside the integral, allowing us to end up with a form that we know how to solve.

Examples

∫xsin(x)dx
Let u(x)=x and v’(x)=sin(x)
Then we need to find u’(x) and v(x), so u’(x) = d/dx(x) = 1
Next, we need to find v(x), which we do by doing ∫v’(x)dx, or in this case, ∫sin(x)dx = -cos(x)
Now that we have u(x), v(x), u’(x), and v’(x), we can plug everything in, resulting in the following:
∫xsin(x)dx = x*(-cos(x)) – ∫1*(-cos(x))dx
= -xcos(x)+∫cos(x)dx
= -xcos(x)+sin(x)+C

In general, we will want to treat the polynomial as u(x) and the other team as v(x) where possible. In addition, Integration by Parts can be done multiple times. Suppose we have the equation

∫x³e^xdx
Then we will let u(x) = x³ and v’(x) = e^x, then u’(x) = 3x² and v(x)=e^x, and we find that
∫x³e^xdx = x³e^x – ∫3x²e^xdx
Then we can do the same procedure with the slightly easier integral ∫3x²e^xdx
Let u(x) = 3x² and v’(x) = e^x, then u’(x) = 6x and v(x) = e^x
Then ∫x³e^xdx = x³e^x – ∫3x²e^xdx = x³e^x – (3x²e^x – ∫6xe^xdx)
And then with ∫6xe^xdx, we split it one last time into u(x) = 6x, v’(x) = e^x, and therefore u’(x) = 6 and v(x) = e^x
Then we can substitute in one last time in order to get ∫6xe^xdx
∫x³e^xdx = x³e^x – ∫3x²e^xdx = x³e^x – (3x²e^x – (6xe^x – ∫6e^xdx)) =
x³e^x – 3x²e^x + 6xe^x – 6e^x + C. Whew!

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