Integration by Parts of Definite Integrals

Integration by Parts of Definite Integrals

Suppose that we have a function that can integrated by parts, but we want to evaluate the integral at some lower bound $a$ and some upper bound $b$. Then it follows that:

(1)
\begin{align} \int_a^b u \: dv = uv \: \rvert_{a}^{b} - \int_a^b v \: du \end{align}

Now let's look at some examples of integration by parts of definite integrals

Example 1

Evaluate the integral $\int_{0}^{\pi} x \cos 3x \: dx$.

First let $u = x$ and $dv = \cos 3x \: dx$. Therefore, $du = dx$ and $v = \frac{1}{3} \sin 3x$. Substituting these values into the integration by parts formula we obtain that:

(2)
\begin{align} \quad \int_{0}^{\pi} x \cos 3x \: dx = \frac{1}{3} u \sin 3x \: \rvert_{0}^{\pi} - \int_{0}^{\pi} \frac{1}{3} \sin 3x \: dx \\ \quad \int_{0}^{\pi} x \cos 3x \: dx = \frac{1}{3} u \sin 3x \: \rvert_{0}^{\pi} + \frac{1}{9} \cos 3x \rvert_{0}{\pi} \\ \quad \int_{0}^{\pi} x \cos 3x \: dx = \frac{1}{3} \left(\pi \cdot 0 - 0 \cdot 0 \right ) + \frac{1}{9} \left ( -1 - 1 \right) \\ \quad \int_{0}^{\pi} x \cos 3x \: dx = -\frac{2}{9} \end{align}