Mathematics, 02.10.2019 08:00, sky724

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

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answered: 030702princessjs

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

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals[Indefinite Integrals] Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

Step-by-step explanation:

Step 1: Define

Identify

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

Step 2: Integrate Pt. 1

Identify variables for integration by parts using LIPET.

Set u: $\displaystyle u = \sin^{-1}(x)$ [u] Arctrig Differentiation: $\displaystyle du = \frac{1}{\sqrt{1 - x^2}} \ dx$ Set dv: $\displaystyle dv = dx$ [dv] Integration Rule [Reverse Power Rule]: $\displaystyle v = x$

Step 3: Integrate Pt. 2

[Integral] Integration by Parts: $\displaystyle \int {\sin^{-1}(x)} \, dx = x \sin^{-1}(x) - \int {\frac{x}{\sqrt{1 - x^2}}} \, dx$

Step 4: Integrate Pt. 3

Identify variables for u-substitution.

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

Step 5: Integrate Pt. 4

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

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

Unit: Integration

answered: Guest

all teachers

step-by-step explanation:

answered: Guest

Answer\ __o_o__/ lil i don’t kniw