Integration by Parts: The LIATE Rule

Some integrals involve a product of two different types of function, such as \( x e^x \) or \( x^2 \sin x \), and substitution does not help. For these, the go-to method is integration by parts. The LIATE rule makes the hardest part, choosing which function to differentiate, almost automatic.

The Integration by Parts Formula

Integration by parts comes from the product rule. Since \( \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx} \), integrating both sides and rearranging gives:

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

You split the integrand into two parts:

• \( u \): the part you will differentiate to get \( du \).
• \( dv \): the part you will integrate to get \( v \).

The goal is for the new integral \( \int v\,du \) to be easier than the original.

The LIATE Rule for Choosing \( u \)

Choose \( u \) as the function that comes first in this list:

1. L – Logarithmic: \( \ln x \), \( \log x \)
2. I – Inverse trigonometric: \( \sin^{-1}x \), \( \tan^{-1}x \)
3. A – Algebraic: \( x \), \( x^2 \), polynomials
4. T – Trigonometric: \( \sin x \), \( \cos x \)
5. E – Exponential: \( e^x \), \( 2^x \)

The function lower in the list becomes \( dv \). This works because functions near the top become simpler when differentiated, while functions near the bottom are easy to integrate.

Worked Example 1: \( \int x e^x\,dx \)

Step 1: Choose using LIATE. \( x \) is Algebraic and \( e^x \) is Exponential. A comes before E, so:

$$ u = x, \qquad dv = e^x\,dx $$

Step 2: Find \( du \) and \( v \).

$$ du = dx, \qquad v = e^x $$

Step 3: Apply the formula.

$$ \int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C $$

Step 4: Check by differentiating. \( \frac{d}{dx}(x e^x - e^x) = e^x + x e^x - e^x = x e^x \). Correct.

We can write the answer neatly as \( e^x(x - 1) + C \).

Worked Example 2: \( \int \ln x\,dx \)

This looks like a single function, but we can write it as \( \ln x \cdot 1 \).

Step 1: Choose. Logarithmic comes first, so:

$$ u = \ln x, \qquad dv = 1\,dx $$

Step 2:

$$ du = \frac{1}{x}\,dx, \qquad v = x $$

Step 3: Apply the formula.

$$ \int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - \int 1\,dx $$

$$ \int \ln x\,dx = x\ln x - x + C $$

The same "multiply by 1" trick works for \( \int \tan^{-1}x\,dx \) and \( \int \sin^{-1}x\,dx \).

Worked Example 3: Repeated Integration by Parts

Find \( \displaystyle\int x^2 \sin x\,dx \).

First application. Algebraic before Trigonometric:

$$ u = x^2, \quad dv = \sin x\,dx \quad\Rightarrow\quad du = 2x\,dx, \quad v = -\cos x $$

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

Second application on \( \int 2x\cos x\,dx \):

$$ u = 2x, \quad dv = \cos x\,dx \quad\Rightarrow\quad du = 2\,dx, \quad v = \sin x $$

$$ \int 2x\cos x\,dx = 2x\sin x - \int 2\sin x\,dx = 2x\sin x + 2\cos x $$

Combine.

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

Check: differentiating gives \( (-2x\cos x + x^2\sin x) + (2\sin x + 2x\cos x) - 2\sin x = x^2\sin x \). Correct.

Each application lowers the power of \( x \) by one, so \( x^n \) needs \( n \) applications.

Worked Example 4: The Cyclic Case \( \int e^x\sin x\,dx \)

Sometimes the original integral reappears. Call it \( I \).

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

First application. Let \( u = \sin x \), \( dv = e^x\,dx \), so \( du = \cos x\,dx \) and \( v = e^x \):

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

Second application. Let \( u = \cos x \), \( dv = e^x\,dx \), so \( du = -\sin x\,dx \) and \( v = e^x \):

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

Substitute back.

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

Solve for \( I \).

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

Important: keep the same type of choice for \( u \) in both steps (trig both times here). Switching would undo your first step.

The Tabular Method: A Shortcut

For integrals like \( \int x^n f(x)\,dx \), you can make a table: differentiate \( u \) repeatedly until you reach 0, integrate \( dv \) repeatedly, then multiply diagonally with alternating signs \( +, -, +, \dots \). For \( \int x^2\sin x\,dx \), the derivatives are \( x^2, 2x, 2, 0 \) and the integrals are \( -\cos x, -\sin x, \cos x \), giving the same answer as Example 3 in a fraction of the time.

Common Mistakes

• Choosing \( u \) the wrong way round. With \( u = e^x \) and \( dv = x\,dx \), the new integral becomes harder.
• Sign errors, especially when \( v = -\cos x \). Use brackets.
• Forgetting the minus in front of \( \int v\,du \).
• Switching choices in a cyclic integral, which just returns \( I = I \).
• Dropping \( + C \) in the final answer.

Key Takeaways

• Integration by parts uses \( \int u\,dv = uv - \int v\,du \), the product rule in reverse.
• Pick \( u \) using LIATE: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential.
• Powers of \( x \) may need repeated applications; the tabular method speeds this up.
• When the original integral reappears, call it \( I \) and solve algebraically.

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