Integration by Parts of Indefinite Integrals

# Integration by Parts of Indefinite Integrals

We are now going to learn another method apart from U-Substitution in order to integrate functions. First recall the product rule for a function in the form $f(x) = u(x)v(x)$:

(1)
\begin{align} \frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \end{align}

Now suppose that we integrate both sides of the product rule. By the The Fundamental Theorem of Calculus Part 1, it follows that:

(2)
\begin{align} u(x)v(x) = \int u'(x)v(x) \: dx + \int u(x)v'(x) \: dx \\ \int u(x)v'(x) \: dx = u(x)v(x) - \int u'(x)v(x) \: dx \\ \end{align}

Now since $\frac{dv}{dx} = v'(x)$, it follows that $dv = v'(x) \: dx$ and similarly, $du = u'(x) \: dx$, and therefore $\int u(x) \: dv = u(x)v(x) - \int v(x) \: du$, or:

(3)
\begin{align} \int u \: dv = uv - \int v \: du \end{align}

Our general goal with integration by parts is to take a rather difficult integral, $\int u \: dv$ and form a simpler-to-evaluate integral, $\int v \: du$. Thus, in general, we generally want to choose a function $u(x)$ that when differentiated is "simpler".

We will now look at some examples of integration by parts.

## Example 1

Evaluate the following integral: $\int xe^x \: dx$.

Let $u = x$, and let $e^x = dv$. We must figure out what our $dx$ is and what our $v$ is. We will summarize this information in the following table:

 $u = x$ $du = dx$ $v = e^x$ $dv = e^x$

We can now substitute these into our equality $\int u \: dv = uv - \int v \: du$. We thus get:

(4)
\begin{align} \int u \: dv = uv - \int v \: du \\ \int x e^x \: dx = xe^x - \int e^x \: dx \\ = xe^x - e^x + C \end{align}

## Example 2

Evaluate the following integral: $\int x \sin 2x \: dx$.

Let $u = x$. It thus follows that $dv = \sin 2x dx$. Hence $du = dx$ and $v = -\frac{1}{2} \cos 2x$. We can thus apply the rule for integration by parts:

(5)
\begin{align} \int x \sin 2x \: dx = -\frac{1}{2}x \cos 2x - \int -\frac{1}{2} \cos 2x \: dx \\ = -\frac{1}{2}x \cos 2x +\frac{1}{2} \int \cos 2x \: dx \\ =-\frac{1}{2}x \cos 2x + \frac{1}{2} \cdot \frac{1}{2} \sin 2x \\ = -\frac{1}{2}x \cos 2x + \frac{1}{4} \sin 2x + C \\ \end{align}

## Example 3

Evaluate the following integral: $\int e^x \cos x \: dx$.

For this integral, let $u = e^x$ and let $dv = \cos x dx$. It thus follows that $du = e^x \: dx$ and $v = \sin x$. Now let's apply our integration by parts identity.

(6)
\begin{align} \int e^x \cos x \: dx = e^x \sin x - \int e^x \sin x \: dx \end{align}

We're now going to use integration by parts once again. Let's keep $u = e^x$, but now let's introduce $p$ and let $p = \sin x dx$. It follows that $du = e^x \: dx$ and $p = -\cos x$. Making the substitution we obtain:

(7)
\begin{align} \int e^x \cos x \: dx = e^x \sin x - [ - e^x \cos x - \int -e^x \cos x \: dx ] \\ \int e^x \cos x \: dx = e^x \sin x + e^x \cos x \\ 2 \int e^x \cos x \: dx = e^x (\sin x + \cos x ) \\ \int e^x \cos x \: dx = \frac{e^x (\sin x + \cos x )}{2} \\ \end{align}

## Example 4

Evaluate the following integral: $\int \tan ^{-1} x \: dx$

Let $u = \tan ^{-1} x$ and $dv = dx$. Therefore $du = \frac{1}{1 + x^2} dx$ and $v = x$. Thus it follows by the integration by parts identity that:

(8)
\begin{align} \int \tan ^{-1} x \: dx = \frac{x^2 tan^{-1} x}{2} - \int \frac{x}{1 + x^2} \: dx \end{align}

We can now use substitution for the remaining integral. Let $p = 1 + x^2$. Then $dp = 2x \: dx$, so we can make the following substitution:

(9)
\begin{align} \int \tan ^{-1} x \: dx = x tan^{-1} x - \int \frac{1}{2} \cdot \frac{1}{p} \: dx \\ \int \tan ^{-1} x \: dx = x tan^{-1} x - \frac{1}{2} \int \frac{1}{p} \: dx \\ \int \tan ^{-1} x \: dx = x tan^{-1} x - \frac{1}{2} \ln p \\ \int \tan ^{-1} x \: dx = x tan^{-1} x - \frac{1}{2} \ln (1 + x^2) \\ \int \tan ^{-1} x \: dx = x tan^{-1} x - \ln \sqrt{(1 + x^2)} \\ \end{align}