- #1

- 47

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## Homework Statement

[tex]\int[/tex][tex]\frac{dx}{1-tan^2(x)}[/tex]

## Homework Equations

n/a

## The Attempt at a Solution

Here's what I've tried:

u=tanx

x=arctanu

dx=[tex]\frac{du}{1+u^2}[/tex]

[tex]\int[/tex][tex]\frac{du}{(1+u^2)(1-u^2)}[/tex]

I then used partial fractions to get:

[tex]\int[/tex][tex]\frac{0.5}{1-u^2}[/tex]+[tex]\frac{0.5}{1+u^2}[/tex]du

[tex]\frac{1}{2}[/tex][tex]\int[/tex][tex]\frac{du}{1+u^2}[/tex]+[tex]\frac{1}{2}[/tex][tex]\int\frac{du}{1-u^2}[/tex]

The left side is simply arctanu while I used trig substitution to solve the right side

[tex]\frac{1}{2}[/tex]arctanu-[tex]\frac{1}{2}[/tex][tex]\int[/tex]csc[tex]\theta[/tex]d[tex]\theta[/tex]

[tex]\frac{1}{2}[/tex]arctan(tanx)+[tex]\frac{1}{2}[/tex]ln(csc[tex]\theta[/tex]+cot[tex]\theta[/tex]) + C

[tex]\frac{1}{2}[/tex]x + [tex]\frac{1}{2}[/tex]ln([tex]\frac{1}{\sqrt{1-u^2}}[/tex]+[tex]\frac{u}{\sqrt{1-u^2}}[/tex]) + C

With more re-substitution, I end up with:

[tex]\frac{1}{2}[/tex]x+[tex]\frac{1}{2}[/tex]ln(tanx+1)-[tex]\frac{1}{2}[/tex]ln([tex]\sqrt{1-tan(x)^2}[/tex]+C

For some reason, this is incorrect because when I try to take the derivative of that, I do not end up with [tex]\frac{1}{1-tan(x)^2}[/tex]

Is there something I'm doing wrong? What should I do?