Mathematics, 14.01.2020 22:31 nanagardiner08

Use the table of integrals, or a computer or calculator with symbolic integration capabilities, to find the indefinite integral.
∫√x^2+15 dx.

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Answer from: Jasminanai42

$\displaystyle \int {\sqrt{x^2 + 15}} \, dx = \frac{15 \ln \big| \sqrt{x^2 + 15} + x \big| + x\sqrt{x^2 + 15}}{2} + C$

General Formulas and Concepts:

Pre-Calculus

Trigonometric Identities

[Right Triangles Only] Pythagorean Theorem: a² + b² = c²

a is a leg b is another leg c is the hypotenuse

Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals[Indefinite Integrals] Integration Constant C

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

U-Substitution

Trigonometric Substitution

Reduction Formula: $\displaystyle \int {\sec^n (u)} \, du = \frac{n - 2}{n - 1} \int {\sec^{n - 2} (u)} \, du + \frac{\sec^{n - 2} (u) \tan (u)}{n - 1}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int {\sqrt{x^2 + 15}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for trigonometric substitution.

Set x: $\displaystyle x = \sqrt{15} \tan \theta$ [x] Differentiate [Trigonometric Differentiation, Derivative Property]: $\displaystyle dx = \sqrt{15} \sec^2 (\theta) \ d\theta$ [x] Rewrite: $\displaystyle \theta = \arctan \bigg( \frac{x}{\sqrt{15}} \bigg)$

Step 3: Integrate Pt. 2

[Integral] Trigonometric Substitution: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = \int {\sqrt{15} \sec^2 (\theta) \sqrt{(\sqrt{15} \tan \theta)^2 + 15}} \, d\theta$ [Integrand] Simplify: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = \int {\sqrt{15} \sec^2 (\theta) \sqrt{15 \tan^2 (\theta) + 15}} \, d\theta$ [Integrand] Factor: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = \int {\sqrt{15} \sec^2 (\theta) \sqrt{15 (\tan^2 (\theta) + 1)}} \, d\theta$ [Integrand] Rewrite [Trigonometric Identities]: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = \int {\sqrt{15} \sec^2 (\theta) \sqrt{15 \sec^2 (\theta)}} \, d\theta$ [Integral] Simplify: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = \int {15 \sec^3 \theta} \, d\theta$ [Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = 15 \int {\sec^3 \theta} \, d\theta$ [Integral] Reduction Formula: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = 15 \bigg( \frac{1}{2} \int {\sec (\theta)} \, d\theta + \frac{\sec (\theta) \tan (\theta)}{2} \bigg)$ [Integral] Trigonometric Integration: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = 15 \bigg( \frac{\ln \big| \tan (\theta) + \sec (\theta) \big|}{2} + \frac{\sec (\theta) \tan (\theta)}{2} \bigg)$ Simplify: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = 15 \bigg( \frac{\ln \big| \tan (\theta) + \sec (\theta) \big| + \sec (\theta) \tan (\theta)}{2} \bigg)$ [θ] Back-Substitute: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = 15 \bigg( \frac{\ln \bigg| \tan \bigg( \arctan \big( \frac{x}{\sqrt{15}} \big) \bigg) + \sec \bigg( \arctan \big( \frac{x}{\sqrt{15}} \big) \bigg) \bigg| + \sec (\theta) \tan (\theta)}{2} \bigg)$ Simplify: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = 15 \bigg( \frac{\ln \bigg| \sqrt{\frac{x^2}{15} + 1} +\frac{x}{\sqrt{15}} \bigg| + \sec (\theta) \tan (\theta)}{2} \bigg)$ [θ] Back-Substitute: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = 15 \bigg( \frac{\ln \bigg| \sqrt{\frac{x^2}{15} + 1} +\frac{x}{\sqrt{15}} \bigg| + \sec \bigg( \arctan \big( \frac{x}{\sqrt{15}} \big) \bigg) \tan \bigg( \arctan \big( \frac{x}{\sqrt{15}} \big) \bigg)}{2} \bigg)$ Simplify: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = 15 \bigg( \frac{\ln \bigg| \sqrt{\frac{x^2}{15} + 1} +\frac{x}{\sqrt{15}} \bigg| + x\sqrt{15} \sqrt{\frac{x^2}{15} + 1}}{2} \bigg) + C$ Simplify: $\displaystyle \int {\sqrt{x^2 + 15}} \, dx = \frac{15 \ln \big| \sqrt{x^2 + 15} + x \big| + x\sqrt{x^2 + 15}}{2} + C$

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

Unit: Integration

Learn more on how to prove the reduction formula used: link