Solved Math Exams

Lesson

Understanding Integration by Parts

Some integrals can't be solved with basic rules or u-substitution alone. When you see a product of two different types of functions—like $x \cdot e^x$ or $x^2 \cdot \sin x$—there's no direct formula to apply. The pieces don't fit together nicely for substitution either.

Integration by parts is the technique for these situations. It's essentially the reverse of the product rule for derivatives. Just like the product rule breaks apart the derivative of a product, integration by parts breaks apart the integral of a product into more manageable pieces.

Key Variables

$u$ — the part you'll differentiate (pick strategically!)
$dv$ — the part you'll integrate
$du$ — the derivative of $u$
$v$ — the antiderivative of $dv$

The Formula

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

In words: The integral of $u \, dv$ equals $u$ times $v$, minus the integral of $v \, du$.

The goal is to pick $u$ and $dv$ so that $\int v \, du$ is easier than what you started with.

How to Choose $u$: The LIATE Rule

When deciding which part of your integrand should be $u$, use the LIATE priority order:

Priority	Type	Examples
1st	Logarithmic	$\ln x$, $\log x$
2nd	Inverse trig	$\arctan x$, $\arcsin x$
3rd	Algebraic	$x$, $x^2$, $x^3$, polynomials
4th	Trigonometric	$\sin x$, $\cos x$
5th	Exponential	$e^x$, $2^x$

Why this order? Functions higher on the list get simpler when you differentiate them, while functions lower on the list stay manageable when you integrate them.

Example 1: Basic Integration by Parts

Problem: Find $\int x e^x \, dx$

Step 1: Identify $u$ and $dv$ using LIATE

We have a product of $x$ (algebraic) and $e^x$ (exponential).

By LIATE: Algebraic comes before Exponential, so:

$u = x$ → $du = dx$
$dv = e^x \, dx$ → $v = e^x$

Step 2: Apply the formula

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

$$\int x e^x \, dx = x \cdot e^x - \int e^x \, dx$$

Step 3: Solve the remaining integral

$$= x e^x - e^x + C$$

Step 4: Factor if possible

$$= \boxed{e^x(x - 1) + C}$$

Example 2: The Tabular Method (Repeated Integration by Parts Shortcut)

When you have a polynomial times an exponential or trig function, you'd normally need to apply integration by parts multiple times. The tabular method is a shortcut that handles all the repetitions at once.

Problem: Find $\int x^3 e^x \, dx$

Step 1: Set up the table

Create two columns:

Left column (D): Repeatedly differentiate $u$ until you get 0
Right column (I): Repeatedly integrate $dv$

By LIATE: $u = x^3$ (algebraic), $dv = e^x \, dx$ (exponential)

D (differentiate)	I (integrate)
$x^3$	$e^x$
$3x^2$	$e^x$
$6x$	$e^x$
$6$	$e^x$
$0$	$e^x$

Step 2: Connect with alternating signs

Draw diagonal arrows from each D entry to the I entry one row below. Alternate signs: $+$, $-$, $+$, $-$, ...

$$\int x^3 e^x \, dx =$$

$$(+)(x^3)(e^x)$$

$$+ (-)(3x^2)(e^x)$$

$$+ (+)(6x)(e^x)$$

$$+ (-)(6)(e^x) + C$$

Step 3: Simplify

$$= x^3 e^x - 3x^2 e^x$$

$$+ 6x e^x - 6e^x + C$$

$$= \boxed{e^x(x^3 - 3x^2 + 6x - 6) + C}$$

Why this works: Each diagonal multiplication represents one application of integration by parts. The table just organizes all the steps at once!

Example 3: The Cycling Technique

Sometimes integration by parts doesn't simplify the integral—it just transforms it into a similar integral. When this happens twice and you get back to your original integral, you can solve algebraically!

Problem: Find $\int e^x \sin x \, dx$

Let's call our integral $I$ to keep things clean:

$$I = \int e^x \sin x \, dx$$

Step 1: First application of IBP

Let $u = \sin x$, $dv = e^x \, dx$

Then $du = \cos x \, dx$, $v = e^x$

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

We still have an integral of the same type. Keep going!

Step 2: Second application of IBP

For $\int e^x \cos x \, dx$:

Let $u = \cos x$, $dv = e^x \, dx$

Then $du = -\sin x \, dx$, $v = e^x$

$$\int e^x \cos x \, dx = e^x \cos x + I$$

(The $+I$ comes from the double negative: $-\int e^x(-\sin x)\,dx = +I$)

Step 3: Substitute back

Substitute into our equation from Step 1:

$$I = e^x \sin x - (e^x \cos x + I)$$

$$I = e^x \sin x - e^x \cos x - I$$

Step 4: Solve for $I$

Add $I$ to both sides:

$$2I = e^x \sin x - e^x \cos x$$

$$I = \frac{e^x(\sin x - \cos x)}{2}$$

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

Key insight: When IBP "cycles" back to the original integral, treat it as a variable and solve algebraically!

Common Mistakes and Misunderstandings

❌ Mistake: Choosing $u$ and $dv$ poorly

Wrong approach: For $\int x e^x \, dx$, letting $u = e^x$ and $dv = x \, dx$

This gives $du = e^x \, dx$ and $v = \frac{x^2}{2}$, leading to:

$$\int x e^x \, dx = \frac{x^2}{2} e^x - \int \frac{x^2}{2} e^x \, dx$$

Why it's wrong: The new integral is harder than the original! We went from $x$ to $x^2$.

Correct: Use LIATE. Algebraic ($x$) comes before Exponential ($e^x$), so $u = x$.

❌ Mistake: Forgetting the constant of integration during cycling

Wrong:

$$2 \int e^x \sin x \, dx = e^x \sin x - e^x \cos x$$

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

Why it's wrong: Even though we're doing algebra with integrals, the final answer still needs $+C$.

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

❌ Mistake: Switching the choice pattern mid-problem in cycling

Wrong: For $\int e^x \sin x \, dx$, choosing $u = \sin x$ first, then $u = e^x$ second.

Why it's wrong: If you switch which function type is $u$ halfway through, you'll undo your first integration by parts and end up with $0 = 0$ (true but useless!).

Correct: Stay consistent. If you start with $u = \sin x$ (trig), keep picking trig functions as $u$ throughout the problem.

❌ Mistake: Using tabular method when it doesn't apply

Wrong: Trying to use the tabular method for $\int \ln x \, dx$

Why it's wrong: The tabular method requires one function that eventually differentiates to 0 (like polynomials). $\ln x$ differentiates to $\frac{1}{x}$, then $-\frac{1}{x^2}$, and never reaches 0.

Correct: Use standard integration by parts: $u = \ln x$, $dv = dx$

Formulas & Reference

Integration by Parts Formula

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

Transforms the integral of a product into a (hopefully simpler) form. Choose u and dv strategically using the LIATE rule.

Variables:

$u$:: the part you differentiate (choose using LIATE priority)
$dv$:: the part you integrate
$du$:: the derivative of u
$v$:: the antiderivative of dv

LIATE Priority Rule

\text{L} \to \text{I} \to \text{A} \to \text{T} \to \text{E}

Memory aid for choosing u in integration by parts. Pick u from the highest priority category present: Logarithmic → Inverse trig → Algebraic → Trigonometric → Exponential.