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2 years ago

Triple the amount of x

Mathematics

2 answers:

Margarita [4]2 years ago

7 0

Answer:

Step-by-step explanation:

Vinvika [58]2 years ago

4 0

Answer:

Triple the amount of x3

Step-by-step explanation:

TRIBLE = 3

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Help with part b ! - please see attachment

vfiekz [6]

Answer:

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

General Formulas and Concepts:
Calculus

Differentiation

Derivatives
Derivative Notation

Integration

Integrals

Integration Method: U-Substitution and U-Solve

Step-by-step explanation:

Step 1: Define

Identify given.

$\displaystyle \int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u:
$\displaystyle u = \sec x$
[u] Apply Trigonometric Differentiation:
$\displaystyle du = \sec x \tan x \ dx$
[du] Rewrite [U-Solve]:
$\displaystyle dx = \cos x \cot x \ du$

Step 3: Integrate Pt. 2

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