Integration by Parts - Study Notes (from Transcript)

Approach to integration by parts

• Problem type: integrals that are products of two functions (e.g., x * e^x) can be tricky to integrate directly.
• Initial checks:
  • See if you can simplify by algebra (combine or reduce the integrand). Often not possible for product examples like x e^x.
  • Try substitution (u-substitution). Try different choices for u. If you pick u = x, then du = dx, which may not match the remaining integrand. If you pick u = x^2, you might get a du that helps, but here it doesn’t.
• When standard tricks fail, apply integration by parts (IBP).
• Core idea: integration by parts rewrites the integral in terms of another integral that is hopefully easier to compute.
• The caveat: you must choose u and dv yourself; the choice is the key to making the integral solvable.
• Preview of a practical trick (LIATE): a heuristic to pick u quickly (described below).
• Practical takeaway: if your first choice for u and dv doesn’t help, try a different one. Practice improves your instinct for good choices.
• Before starting: recap of what you will do with IBP — define u, dv; compute du, v; plug into the formula; simplify; add +C for indefinite integrals.
• Check your result: differentiate the final answer to verify you recover the original integrand.

The integration by parts formula and its meaning

• General formula: $\int u\, dv = uv - \int v\, du$
• What each symbol means:
  • du is the derivative of u times dx; v is the antiderivative of dv.
  • dv must include the differential dx (i.e., dv contains dx).
• Why use IBP: often, the new integral (\int v\, du) is easier to compute than the original.
• The objective: convert a hard integral into a combination of simpler pieces.
• Note on constants: for indefinite integrals, you must include the constant of integration at the end, +C.
• A quick note on derivation: this formula can be derived from the product rule: (\frac{d}{dx}[u v] = u' v + u v'). Integrating both sides leads to the IBP identity.
• There exists a more cumbersome form using f(x), g(x), f'(x), g'(x); the neat, common form comes directly from the product rule and is easier to use.

LIATE trick (the practical u selection rule)

• LIATE stands for:
  • L = Logarithmic: logs like (\ln x) or (\log x)
  • I = Inverse trig: arctan, arcsin, arccos, etc.
  • A = Algebraic: polynomials, powers of x (e.g., (x^2), (x^3))
  • T = Trigonometric: sin, cos, etc.
  • E = Exponential: (e^x), etc.
• Rule of thumb: choose for u the factor that appears earlier in this sequence; differentiate it to get du (which should become simpler); let dv be the other factor (which should be easy to integrate without becoming more complicated).
• The differential dx always accompanies dv (so dv = [something] dx).
• Examples from the transcript:
  • For (\int x e^x\, dx): x is Algebraic (A) and e^x is Exponential (E). Since A comes before E, u = x and dv = e^x dx.
  • For (\int x\sin x\, dx): Algebraic (A) vs Trig (T); A comes before T, so u = x and dv = sin x dx.
  • For (\int x^3 \ln x \; dx): L comes before A, so u = \ln x and dv = x^3 dx.
  • For (\int (x^2 - 1) e^x dx): Algebraic vs Exponential; A wins, so u = x^2 - 1 and dv = e^x dx.
  • For (\int e^x \sin x dx): Exponential vs Trig; E and T can be interchangeable in practice; sometimes you’ll need IBP twice to finish.
  • For {single-term} log or sine integrals (e.g., (\int \ln x \, dx) or (\int \sin x \, dx) treated with IBP): you can think of dx as a separate algebraic term (i.e., use IBP as if the dx is a 1 dx). Then you get a case with log (or inverse trig) as u and 1 dx as dv.
• Special note: If the integrand has only a single term, using the trick still applies by treating the dx as part of the algebraic or a simple dv component.

Worked example: (\int x e^x \; dx) (step-by-step)

• Choose: u = x, dv = e^x dx (as shown by LIATE).
• Compute derivatives/antiderivatives:
  • du = dx
  • v = \int e^x dx = e^x
• Apply IBP: $\int x e^x \; dx = u v - \int v \; du = x e^x - \int e^x dx$
• Integrate the remaining integral: $\int e^x dx = e^x$
• Final form: $\int x e^x \; dx = x e^x - e^x + C = (x-1) e^x + C$
• Quick check: differentiate the result to verify you get the original integrand:
  • (\frac{d}{dx}[(x-1)e^x] = e^x + (x-1)e^x = x e^x).
• Important note: For indefinite integrals, include +C at the end.

Additional examples and practical notes from the transcript

• Example: (\int x \sin x \; dx) using LIATE:
  • u = x (A), dv = sin x dx; v = -cos x; du = dx
  • (\int x \sin x \; dx = -x \cos x - \int -\cos x \, dx = -x \cos x + \sin x + C)
• Example: (\int x^3 \ln x \; dx) uses LIATE with log before algebraic: u = \ln x; dv = x^3 dx; etc. (repeat IBP as needed).
• Example: (\int (x^2 - 1) e^x \; dx) requires IBP twice to complete.
• Example note: If you have an exponential and a trig term, you can switch which is u and which is dv; sometimes you’ll perform the IBP twice to resolve the integral.
• Special case: For exponential and trig, the pairing ET or TE may both work; the choice is flexible and you may need two steps of IBP.
• Example: If the integrand contains log and algebraic parts but the log comes first in LIATE, log is u; the other term is dv.
• For integrals with a mixture of types, you’ll often end up doing IBP more than once.
• Example of a mixed type with a single term (log or sin) treated via IBP by using dx as 1 dx:
  • For (\int \ln x \, dx), using that dx acts like a 1 dx term, you can set u = \ln x and dv = 1 dx, giving du = (1/x) dx, v = x, and the integral becomes $\int \ln x \, dx = x \ln x - \int x \cdot (1/x) dx = x \ln x - x + C.$n

Definite integrals, and a note on derivations and alternate formulas

• Definite integrals: If you have limits, apply IBP with limits to both uv and the remaining integral: evaluate uv from a to b minus the definite integral of v du from a to b.
• Derivation by product rule: You can derive the IBP formula by applying the product rule and integrating; this is a different pathway to the same result (covered in a separate video but acknowledged here).
• Alternative form: There is a more cumbersome version of IBP using f(x), g(x), f'(x), g'(x); the standard compact form is generally neater and easier to use.
• Practical reminder: When dealing with an indefinite integral, remember to include +C; for a definite integral, the limits are applied to both uv and the remaining integral as appropriate.

Tips for mastering integration by parts

• If your first attempt doesn’t simplify the problem, don’t panic; try a different choice of u and dv.
• The more you practice, the better you’ll become at picking u and dv using LIATE or the general rule about which factor becomes simpler on differentiation.
• Always verify with differentiation whether your result matches the original integrand.
• For mixed types, anticipate multiple iterations of IBP (sometimes twice) to resolve the integral.
• Remember the dx belongs with dv; after differentiating u you don’t carry dx in the derivative, but you must keep track of dx when forming du and dv.
• For definite integrals, compute limits carefully for each term (uv evaluated at the bounds and the remaining integral with its limits).

Quick recap cheat sheet

• When to use IBP: for products of two functions where one becomes simpler upon differentiation and the other is easy to integrate.
• Core rule to pick u: differentiate the factor that becomes simpler; let dv be the other factor that you can integrate without becoming more complicated.
• LIATE order (use as a quick guide): Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential.
• Formula: $\int u\, dv = uv - \int v\, du$
• Steps to solve: choose u and dv, compute du and v, apply the formula, simplify, add +C if indefinite, and verify by differentiation.
• Common patterns: several problems require IBP more than once; exponential-trig and algebraic-log combinations follow the same logic.
• Final check: differentiate the result to ensure you recover the original integrand.