integration help needed :) (1 Viewer)

nfreidman

New Member
Joined
Mar 15, 2012
Messages
15
Gender
Female
HSC
2012
how do you integrate:



sin2x/[2+sin2x]


cotx between limits, pi/6 and pi/4

thank you :)
 
Last edited:

nightweaver066

Well-Known Member
Joined
Jul 7, 2010
Messages
1,585
Gender
Male
HSC
2012
For the first one, apply double angle formula to the top so sin2x=2sinxcosx.

Then use the substitution let u = sinx

For the second one, write it as cosx/sinx and notice the numerator is the derivative of the denominator.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top