The tutor shows an essential example of integration by parts.

Back in my July 23 post, I show the formula for integration by parts as

∫uv’=uv-∫vu’

Let’s consider the example

∫xsinxdx

Integration by parts is used to integrate a product of functions. One function needs to be identified as a derivative; it’s the one you integrate. The other function will be differentiated.

In the case of ∫xsinxdx, it’s best to integrate sinxdx, but differentiate x. The reason: differentiating x will get rid of it.

∫xsinxdx = -xcosx – ∫-cosxdx.

which becomes

∫xsinxdx = -xcosx + ∫cosxdx.

Since ∫cosxdx = sinx, we arrive at

∫xsinxdx = -xcosx + sinx + C

The + C part compensates for an unknown constant that was destroyed by the original differentiation.

I’ll be talking more about integral calculus:)

Source:

Larson, Roland and Robert Hostetler. Calculus, third edition.
Toronto: D C Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.