INTEGRATION BY PARTS - EXAMPLE PROBLEMS

 

Example 1 Evaluate the following integral.$$\int x{e}^{6x}dx$$

So, on some level, the problem here is the $x$ that is in front of the exponential. If that wasn’t there we could do the integral. Notice as well that in doing integration by parts anything that we choose for $u$ will be differentiated. So, it seems that choosing $u=x$ will be a good choice since upon differentiating the $x$ will drop out.

Now that we’ve chosen $u$ we know that $dv$ will be everything else that remains. So, here are the choices for $u$ and $dv$ as well as $du$ and $v$.

$$\begin{array}{cccc}u& =x& dv& =e^{6x}dx\\ du& =dx& v& =\int e^{6x}dx=\frac{1}{6}{e}^{6x}\end{array}$$

The integral is then,

$$\begin{array}{cc}\int x{e}^{6x}dx& =\frac{x}{6}{e}^{6x}-\int \frac{1}{6}{e}^{6x}dx\\ & =\frac{x}{6}{e}^{6x}-\frac{1}{36}{e}^{6x}+c\end{array}$$

Once we have done the last integral in the problem we will add in the constant of integration to get our final answer.

Note as well that, as noted above, we know we made made a correct choice for $u$ and $dv$ when we got a new integral that we actually evaluate after applying the integration by parts formula.

Example 2

${\displaystyle \int �\sin 2���}$

Solution

We need to choose ${\displaystyle �}$. In this question we don't have any of the functions suggested in the "priorities" list above.

We could let ${\displaystyle �=�}$ or ${\displaystyle �=\sin 2�}$, but usually only one of them will work. In general, we choose the one that allows ${\displaystyle \frac{��}{��}}$ to be of a simpler form than u.

So for this example, we choose u = x and so ${\displaystyle ��}$ will be the "rest" of the integral, dv = sin 2x dx.

We have ${\displaystyle �=�}$ so ${\displaystyle ��=��}$.

Also ${\displaystyle ��=\sin 2���}$ and integrating gives:

${\displaystyle �=\int \sin 2���}$

${\displaystyle =\frac{-\cos 2�}{2}}$

Substituting these 4 expressions into the integration by parts formula, we get (using color-coding so it's easier to see where things come from):

${\displaystyle \int \underbracket{�}\underbracket{��}}$ ${\displaystyle =\underbracket{�}\underbracket{�}-\int \underbracket{�}\underbracket{��}}$

${\displaystyle \int \boxed{�}\boxed{\sin 2���}=\boxed{�}\boxed{\frac{-\cos 2�}{2}}-\int \boxed{\frac{-\cos 2�}{2}}\boxed{��}}$

${\displaystyle =\frac{-�\cos 2�}{2}+\frac{1}{2}\int \cos 2���}$

${\displaystyle =\frac{-�\cos 2�}{2}+\frac{1}{2}\frac{\sin 2�}{2}+�}$

${\displaystyle =\frac{-�\cos 2�}{2}+\frac{\sin 2�}{4}+�}$