Integration by Parts

This technique is used to compute indefinite integrals.

If two functions f(x) and g(x) are differentiable on the interval [a, b], and their derivatives are continuous, then the following holds: $$ \int f(x)g'(x) \: dx = f(x)g(x)- \int f'(x)g(x) \: dx $$

• The function f(x) is called the finite factor.
• The derivative g′(x) is called the differential factor.

When used appropriately, this formula can greatly simplify the evaluation of an indefinite integral.

Note. Choosing the differential factor wisely is key to applying integration by parts effectively. You should pick the function that makes the resulting integral easier to handle. But this choice isn’t always straightforward.

An Example

Let’s evaluate the following indefinite integral using integration by parts:

$$ \int x \cdot \cos x \: dx $$

The cosine function is the antiderivative of sine.

So we’ll take \( x \) as the finite factor \( f(x) \), and \( \cos x \) as the differential factor \( g'(x) \):

$$ f(x) = x $$

$$ g'(x) = \cos x $$

Integrating g′(x), we get:

$$ g(x) = \sin x $$

Now apply the integration by parts formula:

$$ \int f(x)g'(x) \: dx = f(x)g(x) - \int f'(x)g(x) \: dx $$

$$ \int x \cdot \cos x \: dx = x \cdot \sin x - \int 1 \cdot \sin x \: dx $$

The antiderivative of sine is negative cosine, so we get:

$$ \int x \cdot \cos x \: dx = x \cdot \sin x + \cos x + k $$

We’ve now successfully solved the integral using integration by parts.

Note. The choice of the differential factor is critical. While \( x \) is also the antiderivative of \( \frac{1}{2}x^2 \), had we chosen \( x \) instead of \( \cos x \) as the differential factor, the integral would have been harder to solve: $$ f(x) = \cos x \quad , \quad g'(x) = x $$ So: $$ g(x) = \frac{1}{2}x^2 $$ Now applying the formula: $$ \int f(x)g'(x) \: dx = f(x)g(x) - \int f'(x)g(x) \: dx $$ $$ \int \cos x \cdot x \: dx = \cos(x) \cdot \frac{1}{2}x^2 - \int -\sin x \cdot \frac{1}{2}x^2 \: dx $$ Instead of simplifying, this actually makes the integral more complicated.

Derivation and Explanation

The integration by parts formula is derived from the product rule for differentiation:

$$ [ f(x) \cdot g(x) ]' = f'(x) \cdot g(x) + f(x) \cdot g'(x) $$

Now integrate both sides of the equation:

$$ \int [ f(x) \cdot g(x) ]' \: dx = \int f'(x) \cdot g(x) + f(x) \cdot g'(x) \: dx $$

Using the linearity of integration, the right-hand side becomes a sum of integrals:

$$ \int [ f(x) \cdot g(x) ]' \: dx = \int f'(x) \cdot g(x) \: dx + \int f(x) \cdot g'(x) \: dx $$

Now move the first integral to the left-hand side:

$$ \int [ f(x) \cdot g(x) ]' \: dx - \int f'(x) \cdot g(x) \: dx = \int f(x) \cdot g'(x) \: dx $$

The left-hand side simplifies, since integration and differentiation cancel out:

$$ f(x) \cdot g(x) - \int f'(x) \cdot g(x) \: dx = \int f(x) \cdot g'(x) \: dx $$

This gives us the integration by parts formula:

$$ \int f(x) \cdot g'(x) \: dx = f(x) \cdot g(x) - \int f'(x) \cdot g(x) \: dx $$

This also confirms that the product \( f(x)g(x) \) is an antiderivative of its derivative [f(x)·g(x)]′: $$ \int f(x)g'(x) \: dx = f(x)g(x) - \int f'(x)g(x) \: dx $$ $$ \int f(x)g'(x) \: dx + \int f'(x)g(x) \: dx = f(x)g(x) $$ $$ \int \left(f(x)g'(x) + f'(x)g(x)\right) \: dx = f(x)g(x) $$ Since the sum \( f(x)g'(x) + f'(x)g(x) \) is exactly the derivative of the product, we confirm that: $$ \int [f(x)g(x)]' \: dx = f(x)g(x) $$

And so on.