Integration by Parts

Flashcards covering the concept, formulas, heuristics, and specific examples of the Integration by Parts technique from the Chapter 1, Lecture 7 notes.

Integration by Parts

The integration technique that corresponds to the product rule of differentiation, defined by the formula $\int f(x) \cdot g'(x) \, dx = f(x) \cdot g(x) - \int g(x) \cdot f'(x) \, dx$.

Integration by Parts (Differential Form)

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

Criteria for choosing $u$ and $dv$: (a)

The variable $v$ must be easy to calculate from $dv$, meaning the antiderivative is straightforward.

Criteria for choosing $u$ and $dv$: (b)

The integral $\int v \, du$ must be easier to evaluate than the original integral $\int u \, dv$.

LIATE

A heuristic strategy that proposes a priority order for choosing the variable $u$; it stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential.

Logarithmic (L in LIATE)

The highest priority for choosing the variable $u$ in the integration by parts method.

Inverse Trigonometric (I in LIATE)

The second priority for choosing the variable $u$ in the LIATE strategy.

Algebraic (A in LIATE)

The third priority for choosing the variable $u$ in the LIATE strategy.

Trigonometric (T in LIATE)

The fourth priority for choosing the variable $u$ in the LIATE strategy.

Exponential (E in LIATE)

The lowest priority for choosing the variable $u$, meaning it is often assigned to $dv$.

Definite Integration by Parts Formula

$$\int_{a}^{b} f(x) \cdot g'(x) \, dx = [ f(x) \cdot g(x) ]_{a}^{b} - \int_{a}^{b} g(x) \cdot f'(x) \, dx$$

Integral of $\ln x$

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

Integral of $\sec^{3} x$

$$\int \sec^{3} x \, dx = \frac{1}{2} [ \sec x \cdot \tan x + \ln |\sec x + \tan x| ] + C$$

Trigonometric Identity for $\int \sec^{3} x \, dx$

The formula used during the calculation is $\tan^{2}(x) = \sec^{2}(x) - 1$.

Cyclical Integration (Example $\int e^{x} \cos(x) \, dx$)

A case where integration by parts must be applied twice to return to the original integral expression, allowing it to be solved as an algebraic equation.