yup right
19 Answers
Ankit, I am sure some of the questions you have posted have errors..
either in the answers or the questions....
Can you tell me the source of your questions.. so that I can check them?
i dont see what is wrong
I=0∫2Πecos(2Π-x)cos(2Π-x)sin(2Π-x)
I=same limits n integral of (ecosxcosx(-sinx))
I= -I
nishant you've missed out a minus sign in the power .you won't get I = 0 and one request plz everybody take a bit of time and write properly no . all the powers and subscripts and all . helps ppl like me . who're duds at math . thanks guys.
I= 0∫2pi ecosxcosx(sinx)
I = 0∫2pi ecos(2pi-x)cos(2pi-x)sin(2pi-x)
I = 0∫2pi -ecos(x)cos(x)sin(x)
Thus I=-I
Thus I=0
by taking t=cosx
and then applying limits
cos0=1=cos2Î rite
and please do correct me if im being senseless here
i didnt understand u subash????????
i just did it thinking it to be a indefinite integral. now we have to put limits accordingly na.
let cosx=t -sinx dx=dt
therefore given integral is et t dt
solve the integral by using integration by parts
its now damn easy na dude.