- #1

Sisyphus

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## Homework Statement

Basically, I have to find

[tex]

\int \frac{1}{cosx} dx

[/tex]

by multiplying the integrand by [tex] \frac{cosx}{cosx}

[/tex]

I go through and arrive at a solution, but when I differentiate it,

I get -tan(x)

something's clearly wrong, but I can't see what it is that I'm doing wrong here...

## Homework Equations

[tex]

let u = sin(x)[/tex]

[tex] du = cos(x)dx

[/tex]

## The Attempt at a Solution

[tex]

\int \frac{1}{cosx} dx = \int \frac{cosx}{cos^2x} dx\\

= \int \frac{cosx}{1-sin^2x} dx\\

=\int \frac{du}{1-u^2} \\

=\int \frac{du}{(1-u)*(1+u)} \\

=\frac{1}{2} * \int \frac {1}{1+u} + \frac {1}{1-u} du\\

= \frac{1}{2} * (ln(1-u^2}})

=\frac{1}{2} * (ln(cos^2))

[/tex]

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