Trigonometric Integrals

  • #1

Homework Statement


∫ cosx+sin 2X/sinX


Homework 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)


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
 

Answers and Replies

  • #2
HallsofIvy
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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).
 
  • #3
tiny-tim
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∫ cosx+sin 2X/sinX

Hi afcwestwarrior! :smile:

Hint: one of the standard trigonometric identities …

sin2X = 2 sinX cosX :smile:
 
  • #4
I mean (cos(x)+sin(2x))/ sin(x)

so it would be (cos(x) + 2 sin(x)cos(x))/sin(x)
 
  • #6
so would it be like this
cos(x)/sin(x) - 2 sin(x) cos(x)/ sin (x)= cos(x)/sin(x) - 2cos(x)
 
  • #7
isn't cosX/sinx= to something, i forgot
 
  • #8
tiny-tim
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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).
 
  • #9
I already found the answer thanks once again.
 

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