Integration by parts

Integration by parts is a technique derived from the product rule for differentiation, used to integrate products of functions that cannot be integrated using standard methods. The formula is:

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

Alternatively written as: $\int u(x)v'(x) \, dx = u(x)v(x) - \int u'(x)v(x) \, dx$

Derivation: Starting from the product rule $(uv)' = u'v + uv'$, integrate both sides to obtain $uv = \int u'v , dx + \int uv' , dx$. Rearranging gives the integration by parts formula.

Key Strategy - The LIATE Rule: Choose $u$ using this priority order:

• Logarithmic functions (ln x)
• Inverse trigonometric functions (arctan x, arcsin x)
• Algebraic functions (polynomials: x², x³)
• Trigonometric functions (sin x, cos x)
• Exponential functions (eˣ, e²ˣ)

The function appearing first in LIATE becomes $u$; the remaining factor becomes $dv$.

Mnemonic: "Lions In Africa Take Everything" helps remember LIATE.

Essential Understanding: Integration by parts transforms a difficult integral into (hopefully) a simpler one. Sometimes multiple applications are needed, and occasionally the original integral reappears, allowing algebraic solution.

Definite Integrals: For definite integrals, apply limits after integration: $\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du$

Mastering this technique requires recognizing when to use it and which function to choose as $u$.

Why Integration by Parts Matters: This technique appears extensively in physics, engineering, and economics when modeling complex systems.

Real-World Application 1 - Physics (Rocket Propulsion): When calculating the work done by a variable force over time, engineers often encounter integrals like $\int t e^{-kt} , dt$. Here, thrust decreases exponentially while time increases linearly—a classic integration by parts scenario. The solution helps optimize fuel consumption and trajectory planning.

Real-World Application 2 - Economics (Present Value): Financial analysts calculate present value of continuous income streams using $\int_0^T te^{-rt} , dt$, where $r$ is the discount rate. This models scenarios where earnings grow linearly over time but must be discounted exponentially. Integration by parts provides the exact formula for investment valuations.

Real-World Application 3 - Signal Processing: Electrical engineers analyzing circuit responses encounter integrals like $\int x \sin(\omega x) , dx$. The product of amplitude and oscillation requires integration by parts to determine phase shifts and signal transformations.

Analogy - The "Undo" Process: Think of integration by parts as a strategic exchange. You're trading one difficult integral for another, hoping the new one is easier—like trading complex chess pieces to simplify the board. Sometimes you must make multiple trades (repeated integration by parts) before reaching a solvable position.

Pattern Recognition: Experience teaches you to spot products requiring this technique: polynomial × exponential, polynomial × trigonometric, logarithmic × anything, or inverse trig × polynomial. Each follows predictable patterns once you've practiced the LIATE selection strategy.

Example 1: Evaluate $\int x e^x , dx$

Examiner Note: This fundamental type appears frequently. Show all steps clearly for full marks.

Solution: Using LIATE: $u = x$ (Algebraic), $dv = e^x dx$ (Exponential)

Then: $du = dx$ and $v = e^x$

Applying formula: $\int x e^x \, dx = xe^x - \int e^x \, dx = xe^x - e^x + C = e^x(x-1) + C$

Example 2: Evaluate $\int \ln x , dx$

Examiner Note: Students often don't recognize this needs integration by parts. Remember: ln x is a single function, but treat it as (ln x)(1).

Solution: Rewrite as $\int (\ln x)(1) , dx$

$u = \ln x$ (Logarithmic), $dv = dx$

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

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

Factor: $= x(\ln x - 1) + C$

Example 3: Evaluate $\int_0^{\pi/2} x \sin x , dx$

Solution: $u = x$, $dv = \sin x , dx$

$du = dx$, $v = -\cos x$

$\int_0^{\pi/2} x \sin x \, dx = [-x \cos x]_0^{\pi/2} + \int_0^{\pi/2} \cos x \, dx$ $= [0 - 0] + [\sin x]_0^{\pi/2} = 1$

Common error: Forgetting negative sign from integrating sin x.