Math Help

rtp

integration from 0 to π/2
(2cos^2 x + 2cosx.sinx)dx

p.s: this is not a hw, its a previous exercise in an exam…

thanks ^_^

jajoja

2cosxsinx = sin2x integration => -(cos2x)/2 ==> -(cosn) +1
2cos^2 x = 1 - cos2x or 1+ cos2x dont remember you can check with a calculator
integration ==> n/2 -/+ (sinn)/2

Joe

That doesn't make a difference; we do not solve your exercises. You want help show us your solution, we'll show you were you went wrong.

rtp

ok good
can i do this?

let u = cos x
du = - sinx dx
(2cos^2 x + 2cosx.sinx)dx

-(2u^2 + 2u) du

i cant do this right?

rtp

ok i solved it ... thanks

jajoja

bluewolf wroteok good
can i do this?

let u = cos x
du = - sinx dx
(2cos^2 x + 2cosx.sinx)dx

-(2u^2 + 2u) du

i cant do this right?

i remember yes you can as long as it gives the same answer as i told you (before) so your good to go, make sure you divide by 2 the first line i forgot to do so.

rtp

thanks jajoja

yeah i just remembered it... first i do your part... change the power to 2x and then do a simple integration
u = 2x
du = 2 dx
du/2 = dx
etc..

and i do each one alone ^_^
i think

thanks for the help ^_^