- #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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