Topic 6.11 Notes – Integrating Using Integration by Parts

Integration by Parts

Start with the product rule:

$$\frac{d}{dx}(uv) = u v' + v u'$$

If we reverse that idea and integrate, we get the formula you must know exactly:

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

Everything in this topic flows from that line.

• You split the integrand into two pieces:
  • $u$ → the part you will differentiate
  • $dv$ → the part you will integrate
• Then compute:
  • $du = u' dx$
  • $v = \int dv$

The goal is simple: turn the original integral into a new one that’s easier.

This only makes sense when substitution does not simplify things and the integrand is a product of two different function types.

When Integration by Parts Works Best

Common product types

You’ll usually see:

• Polynomial × exponential
• Polynomial × trig
• Polynomial × $\ln x$
• Exponential × trig
• $x$ × inverse trig

Example:

$$\int x \sin x \, dx$$

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

Then:

• $du = dx$
• $v = -\cos x$

Apply the formula:

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

$$= -x \cos x + \sin x + C$$

Notice the new integral was simpler. That’s what you’re looking for.

Choosing $u$ with LIATE

When you’re unsure, use LIATE:

1. Logarithmic
2. Inverse trig
3. Algebraic
4. Trig
5. Exponential

Pick as $u$ whatever appears earliest in that list.

Why it works: differentiating logarithms, inverse trig, and polynomials usually simplifies them.

Examples:

• $\int x e^{2x} dx$ → $u = x$
• $\int \ln x \, dx$ → rewrite as $\int 1 \cdot \ln x \, dx$, choose $u = \ln x$
• $\int x^3 \cos x \, dx$ → $u = x^3$

LIATE is a guide, not a rule. If your new integral gets worse, switch.

The Method Step by Step

1. Choose $u$ and $dv$
2. Find $du$ and $v$
3. Plug into
   $$\int u\,dv = uv - \int v\,du$$
4. Simplify the new integral
5. Repeat if needed
6. Add $+C$ for indefinite integrals

Quick check: if the algebra gets more complicated, you probably chose poorly.

Special Situations You Must Recognize

Repeated integration by parts

If you have polynomial × trig or polynomial × exponential, you’ll apply IBP multiple times. Each differentiation lowers the polynomial’s degree until it disappears.

This is common on non-calculator sections where they want clean symbolic answers.

When the original integral comes back

With integrals like:

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

After two rounds of IBP, the original integral reappears. You’ll get something like:

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

Then solve algebraically:

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

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

This move shows up regularly in BC questions. Don’t stop when $I$ reappears. Solve for it.

Definite integrals

For definite integrals, you have two clean approaches:

Approach 1
Finish the algebra first, then plug in bounds at the end.

Approach 2
Apply bounds immediately to the $uv$ term and keep them on the remaining integral.

Example setup:

$$\int_0^1 x e^x dx$$

After IBP:

$$= \left[ x e^x \right]_0^1 - \int_0^1 e^x dx$$

Then evaluate normally.

No $+C$. Be careful with $F(b) - F(a)$. Lost negative signs cost points.

Common Mistakes

• Forgetting the minus sign in the formula
• Choosing $u$ and $dv$ in a way that makes the integral harder
• Dropping parentheses when evaluating definite bounds
• Forgetting $+C$
• Stopping before solving for $I$ when it appears on both sides
• Not rewriting $\int \ln x$ as a product