pl help with these

integrate:

1)   (sin4x-2)/1+cos4x

2)  1/sinx(2+cosx)

3)   1/cosx(2+sinx)

 

reply soon

urgent

!!!!!!!!!!!!!!!!!!!!!

Here is the answers/Hints........

 

1) (sin(4x)-2)/(1+cos4x) dx

 

= sin(4x) dx /(1+cos4x) - 2dx/(1+cos(4x))

  

= 2sin(2x)cos(2x) dx /2cos^2(2x) - 21 dx/2cos^2(2x)

 

= tan(2x) - sec^2(2x)

 

 = log{(sec(2x))/2} - {(tan(2x))/2}+C

-------------------------------------------------

-------------------------------------------------

2)  For this,I will give you only hints....you do it......

   

First,put tan(x/2)=t, then we get dx=2/(1+t^2)dt,sinx=2t/(1+t^2),

cosx=(1-t^2)/(1+t^2).substitute these in question,Then do partial fraction,and it will transform in to a standard form.Now integrate using Suitable formula.

3) same as 2nd question