Simple Integral (1 Viewer)

[elseany]

it's a simple trig integral, im not getting the right answer and i cant find a problem in my working :<

can someone please do this with working and post it up

thanks for ya help

http://img523.imageshack.us/img523/9528/integralig6.jpg

 

[webby234]

cos³x = cosx(1 - sin²x)

So the integral = int (cosx/sin² - cosx)
Integral of -cosx = -sinx

For integral of cosx/sin²x
Let u = sinx
du = cosx dx
= Int 1/u²
= -1/u

So integral = -1/sinx - sinx + c

 

[elseany]

yeh thanks, i got that, i must of made a mistaking subbing in the limits >_>

can somebody try subbing in the limits and posting their working? lol

 

[webby234]

lol forgot the limits - couldn't see them properly .

-2 -(- 2.5) = 0.5

 

[elseany]

thankyouu much love webby

i got that answer but when i put them onto this online crap i got -0.5m i know its subtle, but its been bugging me. thanks again

 

[webby234]

Perfectly fine

I would think that you put the limits the wrong way round.