- #1
mrbill
- 6
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Hey guys, I got a little problem for you 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 can't even begin to fathom where they got it from. Here is the work..ill point out the disagreement from the answer key below
[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 that's 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 isn't 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
[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 that's 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 isn't 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