- #1

- 1,753

- 143

## Homework Statement

[tex]

\int {\sec ^3 x\,\,\tan x\,\,dx}

[/tex]

## Homework Equations

[tex]

u = \sec x

[/tex]

This is my guess at u.

## The Attempt at a Solution

[tex]

\frac{{du}}{{dx}} = \sec x\,\,\tan x,\,\,\,dx = \frac{{du}}{{\sec x\,\,\tan x}}

[/tex]

[tex]

\int {\sec ^3 x\,\,\tan x\,\,dx} = \int {u^3 \,\,\tan x\,\,\frac{{du}}{{\sec x\,\,\tan x}}} = \int {u^3 \,\,\tan x\,\,\frac{{du}}{{u\,\,\tan x}} = }

[/tex]

[tex]

\int {u^3 \,\,\,\,\frac{{du}}{u}} = \int {\frac{{u^3 }}{u}} du = \int {u^2 du} = 2u + C =

[/tex]

[tex]

2\sec x + C

[/tex]

But the back of the book says

[tex]

\frac{1}{3}\sec ^3 x + C

[/tex]

Playing around with it on the graphing software, I find that my answer and the book's answer produce different graphs, however, they both seem to have the problem stated in the question as their derivatives. Is one answer more correct than the other?

[tex]

\frac{1}{3}\sec ^3 x + C

[/tex]

[tex]

2\sec x + C

[/tex]