Integration By Parts

The following is an example of using integration by parts (IBP)in a reiterative fashion to solve an integral involving the product of a trigonometric and exponential functions. First let's state the definition of what integration by parts actually is:

Definition of Integration by Parts:

$$\int_{}^{}u \; dv = uv - \int_{}^{}v \; du $$

When determining which part of the integrand is "u" and which part is "dv", consider the following acronym:



L = Logs

I = Inverse Trig

A = Algebraic

T = Trigonometric

E = Exponential

As you travel from left to right along the acronym (from L to E) the first thing you encounter that exists in your particular integrand will be the "u" part of the IBP expression. The left-over part of the integrand will be the "dv"

In a "best-case" scenario, you will successfully integrate after making your substitutions. Occasionally, however, you will run into issues when (after using IBP) you end up with yet another integral that is difficult to evaluate. In such cases, one technique that can be used is to repeatedly implement IBP in a reiterative fashion until you get like integrals on both sides of the equal sign. From there you can manipulate the expression algebraically and solve. Such an example will be demonstrated below:

Example (evaluate the following integral):

$$\int_{}^{}e^{-5t}sin(19.36t) \; dt $$

As we travel from left to right along the acronym LIATE, the 1st thing we come across that exits in our integrand is "T" for "trig". We have the trigonometric sine term in our integrand an as such it will become the "u" term in the IBP expression. The remaining exponential term will become the "dv" term in the IBP expression. We also need to determine du and v.

$$u= sin(19.36t) $$ $$du = 19.36cos(19.36t) $$ $$dv = e^{-5t} $$ $$v = \int_{}^{}e^{-5t} \; dt $$ $$v = \frac{-1}{5}e^{-5t} + A$$

Recalling the definition of IBP:

$$\int_{}^{}u \; dv = uv - \int_{}^{}v \; du $$

...applying to our integral, we get:

$$\int_{}^{}e^{-5t}sin(19.36t) \; dt $$ $$= \frac{-1}{5}e^{-5t} sin(19.36t) - \int_{}^{} \frac{-1}{5}e^{-5t} 19.36cos(19.36t) + A $$ $$= \frac{-1}{5}e^{-5t} sin(19.36t) + 3.872\int_{}^{}e^{-5t}cos(19.36t) + A \qquad (Eqn \; 1)$$

You notice that we again end up in a situation where we have an integral that is difficult to evaluate. We again use LIATE and IBP with the new integral term. While I am somewhat playing fast and loose with constants of integration in order to simply things, definitely don't forget about them. Just recall that every evaluated integral results in a constant and that any constant multiplied, divided, added or subtracted by, to, or from another constant results in a constant:

$$u= cos(19.36t) \qquad \qquad du = -19.36sin(19.36t)$$ $$dv = e^{-5t} \qquad \qquad \qquad v = \int_{}^{}e^{-5t} \; dt$$ $$\qquad \qquad \qquad \qquad \qquad v = \frac{-1}{5}e^{-5t} + A$$

...and plug it into equation #1: $$\int_{}^{}e^{-5t}sin(19.36t) \; dt $$ $$= \frac{-1}{5}e^{-5t} sin(19.36t) + 3.872[ \frac{-1}{5}e^{-5t}cos(19.36t) $$ $$ \qquad - \int_{}^{}(\frac{-1}{5}e^{-5t})(-19.36sin(19.36t)) ] + B$$

After distributing the 3.872 and simplifying: $$\int_{}^{}e^{-5t}sin(19.36t) \; dt = \frac{-1}{5}e^{-5t}sin(19.36t) $$ $$\qquad - 0.7744e^{-5t}cos(19.36t)- 15\int_{}^{}e^{-5t}sin(19.35t) + B$$

Once again we get an integral that is difficult to evaluate. However take note of the fact that we have like integral terms on both sides of the equal sign. Bring the negative 15 term to the left side gives us:

$$16 \int_{}^{}e^{-5t}sin(19.36t) = \frac{-1}{5}e^{-5t}sin(19.36t) - 0.7744e^{-5t}cos(19.36t) + B$$ Dividing through by 16: $$ \int_{}^{}e^{-5t}sin(19.36t) = -0.0125e^{-5t}sin(19.36t) - 0.0484e^{-5t}cos(19.36t) + C$$ Don't forget to carry the constants of integration!!

We have thus successfully evaluated the original given integral.