Trigonometric Integrals

[afcwestwarrior]

1. The problem statement, all variables and given/known data
∫ cosx+sin 2X/sinX


2. Relevant equations
sin^2 x =1/2 (1-cos2x), cos^2=1/2 (1+cos2x)
if cosine is odd, u sin x, cos^2 x=1-sin^x) , if sine is odd u=cosx sin^2x =1-cos^2x)


3. The attempt at a solution
i'm not sure how to start this one because i've never came across a function like this before

[HallsofIvy]

First, clarify. Do you mean cos(x)+ (sin(2x)/sin(x)) or do you mean (cos(x)+ sin(2x))/sin(x)?

Along with formulas for sin2(x) and cos2(x) you also need sin(2x)= 2 sin(x)cos(x).

[tiny-tim]

∫ cosx+sin 2X/sinX

Hi afcwestwarrior! :smile:

Hint: one of the standard trigonometric identities …

sin2X = 2 sinX cosX :smile:

[afcwestwarrior]

I mean (cos(x)+sin(2x))/ sin(x)

so it would be (cos(x) + 2 sin(x)cos(x))/sin(x)

[afcwestwarrior]

Thanks guys

[afcwestwarrior]

so would it be like this
cos(x)/sin(x) - 2 sin(x) cos(x)/ sin (x)= cos(x)/sin(x) - 2cos(x)

[afcwestwarrior]

isn't cosX/sinx= to something, i forgot

[tiny-tim]

so would it be like this
cos(x)/sin(x) - 2 sin(x) cos(x)/ sin (x)= cos(x)/sin(x) - 2cos(x)

That's right! :smile:

And both those are easy to integrate (hint: one's a ln).

[afcwestwarrior]

I already found the answer thanks once again.