Integrals + Trig: Solve sin(2x)/23+cos(x)^2 dx

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I was reviewing my first calc class stuff before starting this second one and came across a problem that i can't 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 can't seem to figure out how to do. Been awhile since I've done this stuff so sorry if its real easy and I am just missing something simple. thanks in advance.
 
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How about let u=23+cos^2x
 
What does the double angle identity \sin{2x} simplify to?
 
Is this a "differential equations" problem?
 
no i don't believe so, its an integral problems that i jus don't understand thought this would be the best place to put it.
 
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.
 
it is the C one u posted IDK how to make it clearer writing all these functions n stuff out with a keyboard
 
This is a link where you can find the basics:

https://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?
 
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SciSteve said:
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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