trig integration problem

[mrbill]

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

\int cot^3 (2x)dx
\int cot^2 (2x) cot (2x)dx
\int (csc^2 (2x) -1)cot (2x)dx
\int (csc^2 (2x) cot (2x) - cot (2x))dx
\int csc^2 (2x) cot (2x)dx - \int cot (2x)dx


u=cot (2x)
du= -2csc^2 (2x)dx
\frac{-1}{2}du=csc^2 (2x)dx
\frac{-1}{2} \int udu
= \frac{1}{2}u^2
= \frac{-1}{4}cot^2 (2x)

u=2x
du=2dx
\frac{1}{2}du = dx
= \frac {1}{2} \ln | \sin (2x) |


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

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

switch them around so the negative isnt sticking out in front
rewrote inverse sin as csc and factored out 1/4
\frac{1}{4}(2 \ln |csc (2x) | - cot^2 (2x))

Heres the problem: the book writes it as:

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

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


mrbill

[KLscilevothma]

I think you're correct.

The answer is \frac{1}{4}(2 \ln |csc (2x) | - cot^2 (2x)) + C but not \frac{1}{4}(2 \ln |csc^2 (2x) | - cot^2 (2x))+ C

[gnome]

It seems odd that your book would put it in that form anyway, when they could have done this:

\frac{1}{4}(ln|csc^2(2x)| - cot^2(2x)) + C


or even this:

\frac{1}{4}(ln|csc^2(2x)| - csc^2(2x)) + C



Probably just a typo in the book.

[modmans2ndcoming]

they are solved by grad students, so you get a bunch of weird ways problems are composed all bunched togehter.

my Calc professor use to have to sit an think why the example is solved the way it is because it made no sense at first some times since a much simpler solution was more conspiquous.