Question

Indefinate integral of sin-1(x? i know you have to use integral by parts: u = sin-1(x v = x

Answer 1

`\displaystyle \int {\sin^(-1)(x)} \, dx = x \sin^(-1)(x) + √(1 - x^2) + C`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• Integrals
• [Indefinite Integrals] Integration Constant C

Integration Rule [Reverse Power Rule]:
`\displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + C`

Integration Property [Multiplied Constant]:
`\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx`

U-Substitution

Integration by Parts:
`\displaystyle \int {u} \, dv = uv - \int {v} \, du`

• [IBP] LIPET: Logs, inverses, Polynomials, Exponentials, Trig

Explanation:

Step 1: Define

Identify

`\displaystyle \int {\sin^(-1)(x)} \, dx`

Step 2: Integrate Pt. 1

Identify variables for integration by parts using LIPET.

1. Set u:
   `\displaystyle u = \sin^(-1)(x)`
2. [u] Arctrig Differentiation:
   `\displaystyle du = (1)/(√(1 - x^2)) \ dx`
3. Set dv:
   `\displaystyle dv = dx`
4. [dv] Integration Rule [Reverse Power Rule]:
   `\displaystyle v = x`

Step 3: Integrate Pt. 2

1. [Integral] Integration by Parts:
   `\displaystyle \int {\sin^(-1)(x)} \, dx = x \sin^(-1)(x) - \int {(x)/(√(1 - x^2))} \, dx`

Step 4: Integrate Pt. 3

Identify variables for u-substitution.

1. Set u:
   `\displaystyle u = 1 - x^2`
2. [u] Basic Power Rule [Derivative Property - Addition/Subtraction]:
   `\displaystyle du = -2x \ dx`

Step 5: Integrate Pt. 4

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {\sin^(-1)(x)} \, dx = x \sin^(-1)(x) + (1)/(2) \int {(2x)/(√(1 - x^2))} \, dx`
2. [Integral] U-Substitution:
   `\displaystyle \int {\sin^(-1)(x)} \, dx = x \sin^(-1)(x) + (1)/(2) \int {(1)/(√(u))} \, du`
3. [Integral] Integration Rule [Reverse Power Rule]:
   `\displaystyle \int {\sin^(-1)(x)} \, dx = x \sin^(-1)(x) + √(u) + C`
4. [u] Back-Substitute:
   `\displaystyle \int {\sin^(-1)(x)} \, dx = x \sin^(-1)(x) + √(1 - x^2) + C`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration