Answer:
(b) ![\displaystyle \int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx = \arctan \big( \sec x \big) + C](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%7B%5Cfrac%7B%5Csec%20x%20%5Ctan%20x%7D%7B1%20%2B%20%5Csec%5E2%20x%7D%7D%20%5C%2C%20dx%20%3D%20%5Carctan%20%5Cbig%28%20%5Csec%20x%20%5Cbig%29%20%2B%20C)
General Formulas and Concepts:
<u>Calculus</u>
Differentiation
- Derivatives
- Derivative Notation
Integration
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](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%7B%5Cfrac%7B%5Csec%20x%20%5Ctan%20x%7D%7B1%20%2B%20%5Csec%5E2%20x%7D%7D%20%5C%2C%20dx)
<u>Step 2: Integrate Pt. 1</u>
<em>Identify variables for u-substitution.</em>
- Set <em>u</em>:
![\displaystyle u = \sec x](https://tex.z-dn.net/?f=%5Cdisplaystyle%20u%20%3D%20%5Csec%20x)
- [<em>u</em>] Apply Trigonometric Differentiation:
![\displaystyle du = \sec x \tan x \ dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20du%20%3D%20%5Csec%20x%20%5Ctan%20x%20%5C%20dx)
- [<em>du</em>] Rewrite [U-Solve]:
![\displaystyle dx = \cos x \cot x \ du](https://tex.z-dn.net/?f=%5Cdisplaystyle%20dx%20%3D%20%5Ccos%20x%20%5Ccot%20x%20%5C%20du)
<u>Step 3: Integrate Pt. 2</u>
- [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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%5Cint%20%7B%5Cfrac%7B%5Csec%20x%20%5Ctan%20x%7D%7B1%20%2B%20%5Csec%5E2%20x%7D%7D%20%5C%2C%20dx%20%26%20%3D%20%5Cint%20%7B%5Cfrac%7B%5Csec%20x%20%5Ctan%20x%20%5Ccos%20x%20%5Ccot%20x%7D%7Bu%5E2%20%2B%201%7D%7D%20%5C%2C%20du%20%5Cleftarrow%20%5C%5C%5Cend%7Baligned%7D)
- [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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%5Cint%20%7B%5Cfrac%7B%5Csec%20x%20%5Ctan%20x%7D%7B1%20%2B%20%5Csec%5E2%20x%7D%7D%20%5C%2C%20dx%20%26%20%3D%20%5Cint%20%7B%5Cfrac%7B%5Csec%20x%20%5Ctan%20x%20%5Ccos%20x%20%5Ccot%20x%7D%7Bu%5E2%20%2B%201%7D%7D%20%5C%2C%20du%20%5C%5C%26%20%3D%20%5Cint%20%7B%5Cfrac%7B1%7D%7Bu%5E2%20%2B%201%7D%7D%20%5C%2C%20du%20%5Cleftarrow%20%5C%5C%5Cend%7Baligned%7D)
- [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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%5Cint%20%7B%5Cfrac%7B%5Csec%20x%20%5Ctan%20x%7D%7B1%20%2B%20%5Csec%5E2%20x%7D%7D%20%5C%2C%20dx%20%26%20%3D%20%5Cint%20%7B%5Cfrac%7B%5Csec%20x%20%5Ctan%20x%20%5Ccos%20x%20%5Ccot%20x%7D%7Bu%5E2%20%2B%201%7D%7D%20%5C%2C%20du%20%5C%5C%26%20%3D%20%5Cint%20%7B%5Cfrac%7B1%7D%7Bu%5E2%20%2B%201%7D%7D%20%5C%2C%20du%20%5C%5C%26%20%3D%20%5Cfrac%7B1%7D%7B1%7D%20%5Carctan%20%5Cbigg%28%20%5Cfrac%7Bu%7D%7B1%7D%20%5Cbigg%29%20%2B%20C%20%5Cleftarrow%20%5C%5C%5Cend%7Baligned%7D)
- 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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%5Cint%20%7B%5Cfrac%7B%5Csec%20x%20%5Ctan%20x%7D%7B1%20%2B%20%5Csec%5E2%20x%7D%7D%20%5C%2C%20dx%20%26%20%3D%20%5Cint%20%7B%5Cfrac%7B%5Csec%20x%20%5Ctan%20x%20%5Ccos%20x%20%5Ccot%20x%7D%7Bu%5E2%20%2B%201%7D%7D%20%5C%2C%20du%20%5C%5C%26%20%3D%20%5Cint%20%7B%5Cfrac%7B1%7D%7Bu%5E2%20%2B%201%7D%7D%20%5C%2C%20du%20%5C%5C%26%20%3D%20%5Cfrac%7B1%7D%7B1%7D%20%5Carctan%20%5Cbigg%28%20%5Cfrac%7Bu%7D%7B1%7D%20%5Cbigg%29%20%2B%20C%20%5C%5C%26%20%3D%20%5Carctan%20u%20%2B%20C%20%5Cleftarrow%20%5C%5C%5Cend%7Baligned%7D)
- [<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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%5Cint%20%7B%5Cfrac%7B%5Csec%20x%20%5Ctan%20x%7D%7B1%20%2B%20%5Csec%5E2%20x%7D%7D%20%5C%2C%20dx%20%26%20%3D%20%5Cint%20%7B%5Cfrac%7B%5Csec%20x%20%5Ctan%20x%20%5Ccos%20x%20%5Ccot%20x%7D%7Bu%5E2%20%2B%201%7D%7D%20%5C%2C%20du%20%5C%5C%26%20%3D%20%5Cint%20%7B%5Cfrac%7B1%7D%7Bu%5E2%20%2B%201%7D%7D%20%5C%2C%20du%20%5C%5C%26%20%3D%20%5Cfrac%7B1%7D%7B1%7D%20%5Carctan%20%5Cbigg%28%20%5Cfrac%7Bu%7D%7B1%7D%20%5Cbigg%29%20%2B%20C%20%5C%5C%26%20%3D%20%5Carctan%20u%20%2B%20C%20%5C%5C%26%20%3D%20%5Cboxed%7B%20%5Carctan%20%5Cbig%28%20%5Csec%20x%20%5Cbig%29%20%2B%20C%20%7D%20%5C%5C%5Cend%7Baligned%7D)
∴ we used substitution to <em>find</em> the indefinite integral.
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Learn more about integration: brainly.com/question/27746468
Learn more about Calculus: brainly.com/question/27746481
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Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
$7.50 * 18.5 hrs = $138.75
$7.50 * 20 hrs = $150
138.75 + 150 = 288.75
She raises the price by ¢0.75
So now it is $8.25 per hour
(I guessed on a calculator and got it right)
Multiply 8.25 by 35 and you get 288.75
Your answer is 35
-GoldenWolfX
Step-by-step explanation:
f(x) = x - 2
f(-1) = -1 -2
= -3
To find the area of a rectangle, we multiply the Base by the Height. In this case the: Base = AB and Height = BC
So to get the area, we multiply AB by BC. (and we know that <em>BC = 5x+5/x+3</em> <em>and BC = [3x+9/2x-4)</em>
So we would do:
x
Note: When multiplying fractions together, we multiply the numerator by the numerator, and the denominator by the denominator.
For example
x
=
Answer:
x
= ![\frac{(5x+5)(3x+9)}{(x+3)(2x-4)}](https://tex.z-dn.net/?f=%5Cfrac%7B%285x%2B5%29%283x%2B9%29%7D%7B%28x%2B3%29%282x-4%29%7D)
=
[<em>Note: </em><em>we are able to factorise some brackets to make the sum easier. For example we can factor out the 5 in (5x+5) to get: 5(x+1) ]</em>
Now lets simplify by multiplying the 5 and 3 together in the numerator:
= ![\frac{15 (x+1)(x+3)}{(x+3)*2(x-2)}](https://tex.z-dn.net/?f=%5Cfrac%7B15%20%28x%2B1%29%28x%2B3%29%7D%7B%28x%2B3%29%2A2%28x-2%29%7D)
If you notice, there is a (x+3) in numerator and the denominator. This means that we can cancel out the (x+3), to get the most simplified expression for the area:
= ![\frac{15(x+1)}{2(x-2)}](https://tex.z-dn.net/?f=%5Cfrac%7B15%28x%2B1%29%7D%7B2%28x-2%29%7D)
Final Simplified Answer:
which is the last option