|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Mar 2008 23:12:08 IST
|
|
|
sin2x dx/(sin^4(x) + cos^4(x))
write sin^4x +cos^4 x =(sin^2(x) +cos^2(x))^2- 2sin^2(x)cos^2(x)=1-2sin^2(x)cos^2(x)
now multiply numerator n dino by 2 so u have 2sin2x.dx/(2-(sin2x)^2) = 2sin2x/1+cos^2(2x) now put cos2x=t
u get it in the form dt/1+t^2
PL DO NOT RATE
|
Nitwit Blubber Odment Tweak
|
this reply: 5 points
(with 1 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
