- #1
noblerare
- 50
- 0
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?