How to integrate cos^3(x)

  • Thread starter kasse
  • Start date
  • #1
kasse
382
1
What is the best way?
 

Answers and Replies

  • #2
rocomath
1,755
1
Idk, you didn't show any work.
 
  • #3
kasse
382
1
I guess I'll have to use an identity. Maybe sin^2(x) = 1-cos^2(x)?

(1-cos^2(x))*sin(x)

u = 1- cos^2(x)

du/dx = -2cos(x)sin(x)


so that


(1-cos^2(x))*sin(x) dx = (-u sin(x)/2cos(x)) du

Doesn't really help, or?
 
  • #4
rocomath
1,755
1
[tex]\int\cos x\cos^2 xdx[/tex]

Use a BASIC trig identity to change the 2nd degree cosine function.
 
  • #5
kasse
382
1
not cosine, sine
 
  • #6
rocomath
1,755
1
not cosine, sine
What are you talking about?
 
  • #7
kasse
382
1
Don't know, I guess I'm too drunk to do maths right now.
 
  • #8
rocomath
1,755
1
Don't know, I guess I'm too drunk to do maths right now.
Try again later :)
 
  • #9
kasse
382
1
Oh, I wrote cos^3 x instead of sin^3 x in the headline. That explains my confusion.

sin^3 x

=

sin^2 x*sin x

=

(1 - cos^2 x)sin x

Then substitution?

u = 1-cos^2 x

du/dx = 2cos x*sin*x

so that

sin^3 x dx = - u / 2cos x

Hm...
 
  • #10
rocomath
1,755
1
distribute the sinx and you'll see your solution
 
  • #11
HallsofIvy
Science Advisor
Homework Helper
43,021
970
Oh, I wrote cos^3 x instead of sin^3 x in the headline. That explains my confusion.

sin^3 x

=

sin^2 x*sin x

=

(1 - cos^2 x)sin x

Then substitution?

u = 1-cos^2 x

du/dx = 2cos x*sin*x

so that

sin^3 x dx = - u / 2cos x

Hm...
So how about just u= cos(x)?

You know what they say "Don't drink and derive"!
 

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