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NISA [10]
2 years ago
15

Help with part b ! - please see attachment

Mathematics
1 answer:
vfiekz [6]2 years ago
3 0

Answer:

(b)  \displaystyle \int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx = \arctan \big( \sec x \big) + C

General Formulas and Concepts:
<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Integration

  • Integrals

Integration Method: U-Substitution and U-Solve

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify given.</em>

<em />\displaystyle \int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

  1. Set <em>u</em>:
    \displaystyle u = \sec x
  2. [<em>u</em>] Apply Trigonometric Differentiation:
    \displaystyle du = \sec x \tan x \ dx
  3. [<em>du</em>] Rewrite [U-Solve]:
    \displaystyle dx = \cos x \cot x \ du

<u>Step 3: Integrate Pt. 2</u>

  1. [Integral] Apply Integration Method [U-Solve]:
    \displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \leftarrow \\\end{aligned}
  2. [Integrand] Simplify:
    \displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \\& = \int {\frac{1}{u^2 + 1}} \, du \leftarrow \\\end{aligned}
  3. [Integral] Apply Arctrigonemtric Integration:
    \displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \\& = \int {\frac{1}{u^2 + 1}} \, du \\& = \frac{1}{1} \arctan \bigg( \frac{u}{1} \bigg) + C \leftarrow \\\end{aligned}
  4. Simplify:
    \displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \\& = \int {\frac{1}{u^2 + 1}} \, du \\& = \frac{1}{1} \arctan \bigg( \frac{u}{1} \bigg) + C \\& = \arctan u + C \leftarrow \\\end{aligned}
  5. [<em>u</em>] Back-substitute:
    \displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \\& = \int {\frac{1}{u^2 + 1}} \, du \\& = \frac{1}{1} \arctan \bigg( \frac{u}{1} \bigg) + C \\& = \arctan u + C \\& = \boxed{ \arctan \big( \sec x \big) + C } \\\end{aligned}

∴ we used substitution to <em>find</em> the indefinite integral.

---

Learn more about integration: brainly.com/question/27746468

Learn more about Calculus: brainly.com/question/27746481

---

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

Unit: Integration

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