AP Calculus – Integration by Parts Made Easy

Integration by parts serves as a method to streamline integrals structured as \(\int f\left(x\right)g\left(x\right) dx\). The method becomes beneficial when function \(f\) can undergo repeated differentiation and function \(g\) can be integrated repeatedly without encountering challenges.

For example,

\(\displaystyle\int x\sin x dx\)

\(\displaystyle\int xe^xdx\)

The formula:

\(\displaystyle\int f\left(x\right)g’\left(x\right)dx=f\left(x\right)g\left(x\right)-\int f’\left(x\right)g\left(x\right)dx\)

or

\(\displaystyle\int udv=uv-\int vdu\)
Integration by Parts

Proof:

The Product Rule says that if \(f\left(x\right)\) and \(g\left(x\right)\) are both differentiable functions, then

\(\displaystyle\frac{d}{dx}\left[f\left(x\right)g\left(x\right)\right]=f\left(x\right)g’\left(x\right)\)

Rewriting this using the notation for indefinite integrals:

\(\displaystyle\int \left[f\left(x\right)g’\left(x\right)+g\left(x\right)f’\left(x\right)\right]dx=f\left(x\right)g\left(x\right)\)

\(\displaystyle\int f\left(x\right)g’\left(x\right)dx+\int g\left(x\right)f’\left(x\right)dx=f\left(x\right)g\left(x\right)\)

Arranging the terms we obtain

\(\displaystyle\int f\left(x\right)g’\left(x\right)dx=f\left(x\right)g\left(x\right)-\int g\left(x\right)f’\left(x\right)\)

Let \(u=f\left(x\right)\) and \(v=g\left(x\right)\):

\(du=f’\left(x\right)dx\) and \(dv=g’\left(x\right)dx\)

Then, by substitution,

\(\displaystyle \int udv=uv-\int vdu\)

Example 1

Find \(\displaystyle\int x\sin xdx\).

Solution

Choose \(f\left(x\right)=x\) and \(g’\left(x\right)=\sin x\):

\(\begin{array}{rcl}\displaystyle\int x\sin xdx&=&f\left(x\right)g\left(x\right)-\int g\left(x\right)f’\left(x\right)dx \\ &=& x\left(-\cos x\right)-\displaystyle\int\left(-\cos x\right)dx \\ &=& -x\cos x+\displaystyle\int\cos xdx \\ &=& -x\cos x+\sin x+C \end{array}\)

Example 2

Evaluate \(\displaystyle\int\ln xdx\).

Solution

Let \(u=\ln x\) and \(dv=dx\):

\(\displaystyle u=\ln x\rightarrow du=\frac{1}{x}dx\)

\(\displaystyle dv=dx\rightarrow v=x\)

Now, let’s integrate by parts:

\(\begin{array}{rcl} \displaystyle\int\ln xdx &=& x\ln x-\int x\cdot\frac{1}{x}dx \\ &=& x\ln x-\displaystyle\int dx \\ &=& x\ln x-x+C \end{array}\)

Example 3

Find \(\displaystyle t^2e^tdt\).

Solution

Let \(u=t^2\) and \(dv=e^tdt\):

\(\begin{array}{rcl} \displaystyle\int t^2e^tdt &=& t^2e^t-2\int te^tdt \\ &=& t^2e^t-2\left(te^t-\displaystyle\int e^tdt\right) \\ &=& t^2e^t-2\left(te^t-e^t+C\right) \\ &=& t^2e^t-2te^t+2e^t+C_1 \end{array}\)

where \(C_1=-2C\).

Example 4

Evaluate \(\displaystyle\int e^x\sin xdx\).

Solution

Let’s first choose \(u=e^x\) and \(dv=\sin x dx\):

\(\displaystyle \int e^x\sin xdx=-e^x\cos x+\int e^x\cos xdx\)

But

\(\displaystyle\int e^x\cos xdx=e^x\sin x-\int e^x\sin xdx\)

Therefore,

\(\displaystyle\int e^x\sin xdx=-e^x\cos x+e^x\sin x-\int e^x\sin xdx\)

Let’s add \(\displaystyle\int e^x\sin xdx\) to both sides of the equation:

\(\displaystyle 2\int e^x\sin xdx=-e^x\cos x+e^x\sin x\)

\(\displaystyle\int e^x\sin xdx=\frac{1}{2}e^x\left(\sin x-\cos x\right)+C\)

Example 5

Calculate \(\displaystyle\int_0^{1}\tan^{-1} xdx\).

Solution

Let \(u=\tan^{-1}x\) and \(dv=dx\):

\(\displaystyle du=\frac{dx}{1+x^2}\qquad v=x\)

Integrate:

\(\begin{array}{rcl} \int_0^1\tan^{-1}xdx &=& x\tan^{-1}x\Bigr|_0^1-\int_0^1\frac{x}{1+x^2}dx \\ &=& \displaystyle 1\cdot \tan^{-1}1-0\cdot \tan^{-1}0-\int_0^1\frac{x}{1+x^2}dx \\ &=& \displaystyle\frac{\pi}{4}-\int_0^1\frac{x}{1+x^2}dx \\ &=& \displaystyle \frac{\pi}{4}-\frac{\ln 2}{2} \end{array}\)

To evaluate \(\displaystyle \int\frac{x}{1+x^2}dx\), use the substitution \(k=1+x^2\).