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Integral using trigonometric sub The answer in the integral is: sin(20) + C 8 4 If x= 2sece then x/2 = sec 0 which is hypotenuse/adjacent. That means hypotenuse=xand adjacent=2. Which left us through Pythagorean theorem with sq rt of x^2-4 as the opposite. Now what will be the value of 0 and the sin(20)to be put in the equation? After that, what will be its derivative so I can make sure I go back the original integral.

Integral using trigonometric sub The answer in the integral is: sin(20) + C 8 4 If x= 2sece then x/2 = sec 0 which is hypotenuse/adjacent. That means hypotenuse=xand adjacent=2. Which left us through Pythagorean theorem with sq rt of x^2-4 as the opposite. Now what will be the value of 0 and the sin(20)to be put in the equation? After that, what will be its derivative so I can make sure I go back the original integral.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Integral using trigonometric sub The answer in the integral is: sin(20) + C 4 If x= 2sece then x/2 = sec 0 which is hypotenuse/adjacent. That means hypotenuse=xand adjacent=2. Which left us through Pythagorean theorem with sqrt of x^2-4 as the opposite. Now what will be the value of 0 and the sin(20)to be put in the equation? After that, what will be its derivative so I can make sure I go back the original integral.
Transcribed Image Text:Integral using trigonometric sub The answer in the integral is: sin(20) + C 4 If x= 2sece then x/2 = sec 0 which is hypotenuse/adjacent. That means hypotenuse=xand adjacent=2. Which left us through Pythagorean theorem with sqrt of x^2-4 as the opposite. Now what will be the value of 0 and the sin(20)to be put in the equation? After that, what will be its derivative so I can make sure I go back the original integral.
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