Posted by Carol on Monday, January 3, 2011 at 8:36pm.
I'm new to this. I didn't know you could reply to the original post? (testing with this reply), otherwise I would not have kept re-posting as a 'new' question.
I just noticed some replies as I was reading other questions. again, my apologies. I feel like I should change my name if I have another question! have a good night.
Carol
Since we did not resolve your question, I am curious where you got this problem.
Is it from a textbook?
Is in an assignment problem?
Just curious, since it obviously not a true statement.
It's actually the end of an integration problem from a college calc text I have from 1971. The problem involved Weierstrass Substitution. The part I posted was after converting from the u-substition to an answer in terms of x.
The orig prob. is
integrate (cos x )/(3 cos^2 x - 5 cos x - 4 sin x cos x ) dx
I'll paste the part I was referring to in a new reply post.
Continued, also this isn't an official assignment. I'm teaching myself calculus. I just love math. I took cal 30 thirty ago but don't remember much.
I use Wolframe Alpha to check my answers.
here's the web paste without the beg. address part, you'll have to copy and paste. look at the bottom just before the graph.
wolframalpha dotcom period slash input/?i=integrate+%28cos+x++%29%2f%283+cos^2+x+-+5+cos+x+-+4+sin+x+cos+x+%29++dx&asynchronous=pod&s=29&incTime=true
I did manage to cut and paste your link, here is what I found:
http://www.wolframalpha.com./input/?i=integrate+%28cos+x++%29%2f%283+cos^2+x+-+5+cos+x+-+4+sin+x+cos+x+%29++dx&asynchronous=pod&s=29&incTime=true
I did notice that the equivalence was between
-sin(x/2)/(2sin(x/2) + cos(x/2)) and
- 1/(cot(x/2) + 2)
which is different from what you had.
as I said it my first reply to an earlier post, since all angles are x/2 , for convenience I will use Ø for x/2
so let's prove
-sinØ/(2sinØ + cosØ) = -1/(cotØ + 2)
RS = -1/(cosØ/sinØ + 2)
= -1/((cosØ + 2sinØ)/sinØ)
= =sinØ/(2sinØ + cosØ) , I multiplied by the reciprocal
= LS
OK then!!!! We got it.
Btw, good for you to work on this all by yourself.
Looks like you have reached quite an advanced stage already.
sorry for the late thank you!!
I lost the post and couldn't find it.
doubt you will see this, and I didn't want to clog up the page with a new post.
you're the greatest!!