Hey guys, I got a little problem for ya involving trig integration. I have listed my work below. My question is....well...the back of the book has a csc^2 (2x) in the absolute value at the end of the problem..and i cant even begin to fathom where they got it from. Here is the work..ill point out the disagreement from the answer key below(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int cot^3 (2x)dx[/tex]

[tex]\int cot^2 (2x) cot (2x)dx[/tex]

[tex]\int (csc^2 (2x) -1)cot (2x)dx[/tex]

[tex]\int (csc^2 (2x) cot (2x) - cot (2x))dx[/tex]

[tex]\int csc^2 (2x) cot (2x)dx - \int cot (2x)dx[/tex]

[tex]u=cot (2x)[/tex]

[tex]du= -2csc^2 (2x)dx[/tex]

[tex]\frac{-1}{2}du=csc^2 (2x)dx[/tex]

[tex]\frac{-1}{2} \int udu[/tex]

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

[tex]= \frac{-1}{4}cot^2 (2x)[/tex]

[tex]u=2x[/tex]

[tex]du=2dx[/tex]

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

[tex]= \frac {1}{2} \ln | \sin (2x) |[/tex]

rewrite and move the negative to an exponent using properties of natural log..its stupid but thats how the textbook has the answer

[tex] \frac{-1}{4} cot^2 (2x) + \frac{1}{2} \ln (sin (2x))^-1[/tex]

switch them around so the negative isnt sticking out in front

rewrote inverse sin as csc and factored out 1/4

[tex]\frac{1}{4}(2 \ln |csc (2x) | - cot^2 (2x))[/tex]

Heres the problem: the book writes it as:

[tex]\frac{1}{4}(2 \ln |csc^2 (2x) | - cot^2 (2x))[/tex]

notice the csc^2 up there...cant figure it out!

mrbill

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# Trig integration problem

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