integration of trig (1 Viewer)

[angmor]

can someone explain to me how

cos x / sin x = ln (sin x) ?

thanks

 

[SoulSearcher]

int. [cos x/sin x] dx
Now let f(x) = sin x
f'(x) = cos x
Therefore,
int. [cos x/sin x] dx
= int. [f'(x)/f(x)] dx, where f(x) = sin x
= ln {f(x)} + C
= ln (sin x) + C

 

[passion89]

I have a question and thought it would be useless starting another thread:

Integrate:
sin[pi(x) / 2]

 

[SoulSearcher]

int. [sin (pi(x)/2)] dx
= -cos(pi(x)/2) / (pi/2)
= -2cos(pi(x)/2) / pi

 

[passion89]

int. [sin (pi(x)/2)] dx
= -cos(pi(x)/2) / (pi/2)
= -2cos(pi(x)/2) / pi

erm I'm still a bit confused...

 

[SoulSearcher]

Ok then, I'll start by finding the derivative of [-2cos(pi(x)/2)]/ pi
y = [-2cos(pi(x)/2)]/ pi
= -2/pi * cos u, where u = pi(x)/2
dy/du = -2/pi * -sin u
= 2/pi sin u
du/dx = pi/2
.'. dy/dx = dy/du * du/dx
= 2/pi sin u * pi/2
= sin u
= sin (pi(x)/2)
From there you can see that the integral of sin (pi(x)/2) is [-2cos(pi(x)/2)]/ pi

Hope that helps

 

[passion89]

Ohhhh right...
thanks!

 

[vanush]

can someone explain to me how

cos x / sin x = ln (sin x) ?

thanks

Do you remember how to differentiate logs?

y'=

f'(x)
-----
f(x)

so for ln (sin x)

it would be

derivative of sin x
----------------------
         sin x

which is equal to

cos x 
------
sin x