Question

Help with part a ! - please see attachment

Answer 1

Answer:

(a)  $\displaystyle \int {\frac{\sec x \tan x}{1 + \sec x}} \, dx = \boxed{ \ln | 1 + \sec x | + C }$

General Formulas and Concepts:
Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle (u + v)' = u' + v'$
Derivative Rule [Basic Power Rule]:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• Integrals

Integration Methods: U-Substitution and U-Solve

Step-by-step explanation:

Step 1: Define

Identify given.

$\displaystyle \int {\frac{\sec x \tan x}{1 + \sec x}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution/u-solve.

1. Set u:
   $\displaystyle u = 1 + \sec x$
2. [u] Differentiate [Derivative Rules, Properties, and Trigonometric Differentiation]:
   $\displaystyle du = \sec x \tan x \ dx$
3. [du] Rewrite [U-Solve]:
   $\displaystyle dx = \cos x \cot x \ du$

Step 3: Integrate Pt. 2

1. [Integral] Apply Integration Method [U-Solve]:
   $\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u}} \, du\end{aligned}$
2. [Integrand] Simplify:
   $\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u}} \, du \\& = \int {\frac{1}{u}} \, du \\\end{aligned}$
3. [Integral] Apply Logarithmic Integration:
   $\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u}} \, du \\& = \int {\frac{1}{u}} \, du \\& = \ln | u | + C \\\end{aligned}$
4. [u] Back-substitute:
   $\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u}} \, du \\& = \int {\frac{1}{u}} \, du \\& = \ln | u | + C \\& = \boxed{ \ln | 1 + \sec x | + C } \\\end{aligned}$

∴ we have used substitution to find the indefinite integral.

---

Learn more about integration: brainly.com/question/27746485

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

---

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

Unit: Integration