Welcome back to class! In Section 7-1, we are diving into one of the most powerful techniques in our integration toolkit: Integration by Parts. If you remember that $u$-substitution is essentially the reverse Chain Rule, you can think of Integration by Parts as the reverse Product Rule.

The Formula

Recall the Product Rule from differentiation: $\frac{d}{dx}[f(x)g(x)] = f(x)g'(x) + g(x)f'(x)$. By integrating both sides and rearranging the terms, we arrive at our formula for this section:

$$ \int u \, dv = uv - \int v \, du $$

The goal here is to trade a difficult integral ($\int u \, dv$) for an easier one ($\int v \, du$).

Strategy: The LIATE Rule

The hardest part of this method is deciding which part of your integrand should be $u$ and which part should be $dv$. If you pick the wrong one, the integral gets more complicated, not simpler (as seen in our class notes where choosing $u=\sin x$ for $\int x \sin x \, dx$ made things messy!).

To choose $u$, we use the hierarchy LIATE. Choose the function type that appears highest on this list to be your $u$:

  • L: Logarithmic functions (e.g., $\ln x$)
  • I: Inverse Trigonometric functions (e.g., $\tan^{-1} x$)
  • A: Algebraic functions (e.g., $x$, $x^2$)
  • T: Trigonometric functions (e.g., $\sin x$, $\cos x$)
  • E: Exponential functions (e.g., $e^x$, $3^x$)

Classic Example: Algebraic $\times$ Trig

Let's look at the example from class: $$\int x \sin x \, dx$$

Using LIATE, A (Algebraic $x$) comes before T (Trig $\sin x$).
So, we let $u = x$ and $dv = \sin x \, dx$.
Then, we differentiate $u$ to get $du = 1 \, dx$, and integrate $dv$ to get $v = -\cos x$.

Plugging this into our formula:

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

The "Invisible" Function

Sometimes you only see one function, like in $\int \ln x \, dx$. How do we use parts? The trick is to let $dv = 1 \, dx$.

Since L comes first in LIATE, we set $u = \ln x$.
$du = \frac{1}{x} \, dx$ and $v = x$.

$$ \text{Result: } x \ln x - x + C $$

The "Boomerang" (Circular Integration)

One of the trickiest examples in our notes is $\int e^x \sin x \, dx$. Since Exponentials and Trig functions cycle when derivatives are taken, you might feel like you are going in circles.

If you apply Integration by Parts twice, you will actually arrive back at the original integral on the right side of your equation!

$$ \int e^x \sin x \, dx = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx $$

Don't panic! Treat the integral as a variable. Add $\int e^x \sin x \, dx$ to both sides, then divide by 2. The final result (derived on page 12 of the notes) is:

$$ \int e^x \sin x \, dx = \frac{1}{2}(e^x \sin x - e^x \cos x) + C $$

Keep practicing the LIATE rule, and don't forget to evaluate your bounds if you are working with a definite integral (like the $\tan^{-1}x$ problem from class). Good luck studying!