Week 01-2: Integration by Parts

Integration by Parts

The rule for integration parts comes from the Product Rule for differentiation.

In the notation for definite integrals, this becomes $$ \int [f'(x)g(x) + f(x)g'(x)] dx = f(x)g(x), $$

or $$ \int f'(x)g(x) dx + \int f(x)g'(x) dx = f(x)g(x). $$

We can use this to create the formula for integration by parts: $$ \boxed{\int f(x)g'(x) dx = f(x)g(x) - \int f'(x)g(x) dx}. \tag*{1} $$

If we let \(u=f(x)\) and \(v=g(x)\), by the Substitution Rule, the formula becomes: $$ \boxed{\int u dv = uv - \int v du}. \tag*{2} $$

Solved Example 1

Find \( \int x \cos x dx \).

1st Method Solution

Choose \( f(x) = x \) and \( g'(x) = \cos x \). We then have \( f'(x) = 1 \) and \( g(x) = \sin x \). Thus, using Formula 1, we have:

\begin{align} \int x \cos x dx &= x \sin x - \int \sin x dx \\ &= x \sin x - (- \cos x) \\ &= \boxed{x \sin x + \cos x + C} \end{align}

Checking by differentiation, we get \( x \cos x \), thus verifying that we reached the correct answer.

2nd Method Solution (using a table)

Let

And so,

\begin{align} \int x \cos x dx &= \int \overbrace{x}^{u} \overbrace{\cos x dx}^{dv} = \overbrace{x}^{u} \sin x - \int \sin x dx \\ &= x \sin x - (- \cos x) \\ &= \boxed{x \sin x + \cos x + C} \end{align}