4. Integration by Parts

Looking at the table of Derivative and Integral Rules, we are missing an integral version of the Product Rule. It is called Integration by Parts which is the topic of this chapter.

a1. Indefinite Integrals

In general, a derivative rule produces an indefinite integral rule: \[ \dfrac{d}{dx}\left[F(x)\right]=f(x) \qquad \text{becomes} \qquad \int f(x)\,dx=F(x)+C \] So the Product Rule \[ \dfrac{d}{dx}[u(x)v(x)]=u(x)\dfrac{dv}{dx}+v(x)\dfrac{du}{dx} \] produces the indefinite integral rule \[ \int \left[u(x)\dfrac{dv}{dx}+v(x)\dfrac{du}{dx}\right]\,dx=u(x)v(x)+C \] This integral rule is not yet in a useful form. A useful form is produced if we split up the integral on the left as the sum of two integrals and take one of them to the other side of the equation:

\[ \int u(x)\dfrac{dv}{dx}\,dx=u(x)v(x)-\int v(x)\dfrac{du}{dx}\,dx \]

Notice that we have dropped the constant of integration since there is an indefinite integral on both sides which (when evaluated) will provide the constant.

This form is more useful because it helps us compute the integral of a product. If the integrand is the product of \(u(x)\) and \(\dfrac{dv}{dx}\), then we can rewrite the integral with the integrand being the product of \(v(x)\) and \(\dfrac{du}{dx}\). Hopefully, the integral \(\displaystyle \int v(x)\dfrac{du}{dx}\,dx\) will be easier to compute.

Frequently, students incorrectly assume:

Wrong! The integral of a product
is the product of the integrals. Wrong!

It is NOT!

By introducing the differentials \[ du=\dfrac{du}{dx}\,dx \qquad \text{and}\qquad dv=\dfrac{dv}{dx}\,dx \] the integration by parts formula may be written in a form which is easier to remember:

\[ \int u\,dv=u\,v-\int v\,du \] where \(du=\dfrac{du}{dx}\,dx\) and \(dv=\dfrac{dv}{dx}\,dx\).

Memorize this!

Notice that the integration by parts formula does not compute the integral but rather transforms it into another integral which is hopefully easier to compute. This should be clarified by the following example:

Compute \(\displaystyle \int x\cos x\,dx.\)

To apply the formula, we need to identify \(u\) and \(dv.\) If we take \(u=x\) and \(dv=\cos x\,dx\), then \(du=dx\) and \(v=\sin x.\)

\[u=x \qquad \Longrightarrow \qquad \dfrac{du}{dx}=1 \qquad \Longrightarrow \qquad du=\dfrac{du}{dx}\,dx=1\,dx=dx\] \[dv=\cos x\,dx=\dfrac{dv}{dx}\,dx \qquad \Longrightarrow \qquad \dfrac{dv}{dx} =\cos x \qquad \Longrightarrow \qquad v=\sin x\]

So: \[\int x\cos x\,dx=u\,v-\int v\,du=x\sin x-\int \sin x\,dx\] Now the final integral is easy to perform: \[ \int x\cos x\,dx=x\sin x-(-\cos x)+C=x\sin x+\cos x+C . \]

We check by differentiating (using the Product Rule). If \(f(x)=x\sin x+\cos x\), then: \[ f'(x)=(\sin x+x\cos x)-\sin x=x\cos x \] which is the integrand we started with.

We usually write \(u\), \(dv\), \(du\) and \(v\) in an array such as: \[\begin{array}{ll} u=x & dv=\cos x\,dx \\ du=dx \quad & v=\sin x \end{array}\] Get in this habit.

Compute the integral \(\displaystyle \int x\sin x\,dx\).

\(\displaystyle \int x\sin x\,dx\,=-x\cos x+\sin x+C\)

Let \[\begin{array}{ll} u=x & dv=\sin x\,dx \\ du=dx \quad & v=-\cos x \end{array}\] Then \[\begin{aligned} \int x\sin x\,dx &=-x\cos x-\int -\cos x\,dx \\ &=-x\cos x+\int \cos x\,dx \\ &=-x\cos x+\sin x+C \end{aligned}\]

We check (using the Product Rule). If \(f(x)=-x\cos x+\sin x\), then: \[ f'(x)=-\cos x+x\sin x+\cos x=x\sin x \] which is the integrand we started with.

Compute the integral \(\displaystyle \int e^x x\,dx\).

The main difficulty is deciding what to take for \(u\) and what to take for \(dv\). Suppose we first try \[\begin{array}{ll} u=e^x & dv=x\,dx \\ du=e^x\,dx \quad & v=\dfrac{x^2}{2} \end{array}\] Then: \[ \int e^x x\,dx=e^x\dfrac{x^2 }{2}-\int e^x\dfrac{x^2 }{2}\,dx \] This is correct but notice that the remaining integral is more complicated than the original integral. So it is better to choose the parts according to: \[\begin{array}{ll} u=x & dv=e^x\,dx \\ du=dx \quad & v=e^x \end{array}\]

\(\displaystyle \int e^x x\,dx=xe^x-e^x+C\)

Let: \[\begin{array}{ll} u=x & dv=e^x\,dx \\ du=dx \quad & v=e^x \end{array}\] Then: \[\begin{aligned} \int e^x x\,dx &=xe^x-\int e^x\,dx \\ &=xe^x-e^x+C \end{aligned}\]

Using the Product Rule, if \(f(x)=xe^x-e^x\), then \[ f'(x)=(e^x+xe^x)-e^x=xe^x \] as desired.

4. Integration by Parts

a1. Indefinite Integrals

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