   Chapter 8.4, Problem 35E

Chapter
Section
Textbook Problem

# Finding an Indefinite Integral In Exercises 19-32, find the indefinite integral. ∫ arcsec 2 x   d x ,   x > 1 2

To determine

To calculate: The solution of the following indefinite integral arcsec2xdx.

Explanation

Given:

The indefinite integral arcsec2xdx.

Formula used:

uvdx=uvdx(dudxvdx)dx

Calculation:

Consider thefollowing indefinite integral;

arcsec2x.1dx

Now, suppose v=1 and u=arcsec2x andapply above formula,

arcsec2x.1dx=arcsec2x1dx(d(arcsec2x)dxxdx)dxarcsec2x.1dx=arcsec2x(x)1x4x21xdxarcsec2x.1dx=xarcsec2x1(2x)21dx

Let’s suppose I=1(2x)21dx.

Then, the above expression become as,

arcsec2x

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