Integration by parts

Now we consider one of the most powerful methods of evaluating the product of two integrable functions.

Let f and g derivable functions in [a,b], we have

(fg)' = f'g + fg'

that is:

fg' = (fg)' − f'g

taking the indefinite integral of both sides, and noticing that ∫ (fg)'dx = fg, we find the integration by parts rule:

∫f(x) ⋅ g'(x) = f(x) ⋅ g(x) − ∫f'(x) ⋅ g(x) dx

which can be applied as well for the definite integral:

\int_{a}^{b} f (x) \cdot g^{'} (x) d x = {[f (x) \cdot g (c)]}_{a}^{b} - \int_{a}^{b} f^{'} (x) \cdot g (x) d x

Example 1. Evaluate

∫ e^x sin x dx

We are able to calculate the primitive of both functions e^x and sin x. We proceed as follows:

∫ e^x sin x dx = e^x sin x − ∫ e^x cos x dx

We perform a second integrations by parts, picking f,g' as in the first choiche:

= e^x sin x − { e^x cos x − ∫ e^x (−sin x) dx}

Let I be the original integra, then we have

I = e^x sin x − e^x cos x − I

By solving the equation we obtain I

2I = e^x sin x − e^x cos x

I = 1/2 e^x (sin x − cos x) + c. ■

Example 2. Evaluate ∫ log x, by parts, by letting f(x) = log x and g'(x) = 1:

∫ log x dx = x log x − ∫x ⋅ 1/x = x log x ⋅ ∫ 1 ⋅ dx = x logx − x + c. ■