   Chapter 4.9, Problem 17E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find the most general antiderivative of the function. (Check your answer by differentiation.)h(θ) = 2 sin θ − sec2 θ

To determine

To find: The most general antiderivative of the function h(θ)=2sinθsec2θ and check the determined antiderivative for the function h(θ)=2sinθsec2θ by differentiation.

Explanation

Given Data:

Write the given function as follows.

h(θ)=2sinθsec2θ

Formula used 1:

The antiderivative function for the function sinθ is cosθ+C .

Here, C is the constant.

The antiderivative function for the function sec2θ is tanθ+C .

Formula used 2:

Write the required differentiation formula to verify the answer as follows.

ddθ(cosθ)=sinθddθ(tanθ)=sec2θddθ(constant)=0

Calculation:

Write the function as follows.

h(θ)=2sinθsec2θ (1)

From the antiderivative function formulae, the antiderivative function for the function in equation (1) is written as follows.

H(θ)=2cosθtanθ+C

Consider the constant C as Cn , since the constant is valid for the interval (nππ2,nπ+π2) .

Therefore the antiderivative is written as follows.

H(θ)=2cosθtanθ+Cn,(nππ2)<Cn<(nπ+π2)

Thus, the most general antiderivative of the function h(θ)=2sinθsec2θ is 2cosθtanθ+Cn,(nππ2)<Cn<(nπ+π2)_

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Study Guide for Stewart's Multivariable Calculus, 8th 