View Full Version : Integrals + Trig
SciSteve
Jan24-08, 08:03 PM
I was reviewing my first calc class stuff before starting this second one and came across a problem that i cant seem to get, its the integral of sin(2x)/23+cos(x)^2 dx, i know most of the rules and thought i had it but the question asks to put in all trig functions in terms of cos which I cant seem to figure out how to do. Been awhile since I've done this stuff so sorry if its real easy and im just missing something simple. thanks in advance.
rock.freak667
Jan24-08, 08:50 PM
How about let u=23+cos^2x
rocomath
Jan24-08, 08:56 PM
What does the double angle identity \sin{2x} simplify to?
HallsofIvy
Jan25-08, 07:29 AM
Is this a "differential equations" problem?
SciSteve
Jan25-08, 01:33 PM
no i dont believe so, its an integral problems that i jus dont understand thought this would be the best place to put it.
coomast
Jan26-08, 02:36 AM
What exactly is the question? Is it:
A=\int \frac{sin(2x)}{23+cos(x^2)}dx
or:
B=\int \left(\frac{sin(2x)}{23}+cos(x^2) \right)dx
or:
C=\int \frac{sin(2x)}{23+cos^2(x)}dx
or:
D=\int \left(\frac{sin(2x)}{23}+cos^2(x) \right)dx
It is unclear what you mean.
Almost certainly C. Though I get your point, careless notation is annoying.
SciSteve
Jan26-08, 11:34 AM
it is the C one u posted IDK how to make it clearer writing all these functions n stuff out with a keyboard
coomast
Jan26-08, 11:46 AM
This is a link where you can find the basics:
http://www.physicsforums.com/misc/howtolatex.pdf
If you click on a formula, the latex code pops up. No worries about the typing, you'll learn it. So the third one is the one you need to tackle. What have you got so far?
Hootenanny
Jan26-08, 11:47 AM
it is the C one u posted IDK how to make it clearer writing all these functions n stuff out with a keyboard
In that case consider re-writing the denominator;
23+cos^2(x) = 23 + \frac{1}{2}\left(1 + \cos(2x)\right) = \frac{1}{2}\left(47+\cos(2x)\right)
Now take the derivative and compare with the numerator.
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