Question

∫pi/2 between 0 of : e^(cos x) sin x dx ? I GIVE MEDALS

Answer 1

let u = cosx
du = -sinx dx

Answer 2

yeah I already did this :) but after ?

Answer 3

after that integrate !
integral -e^u du

Answer 4

yeah but it doesnt work :(

Answer 5

cuz the answer is :e-1= 1,72

Answer 6

but cos of pi/2 = 0 .... :(

Answer 7

this is ur integral right ?
\[\int\limits_{0}^{\frac{ \pi }{2 }} e^\cosx \sin x dx \]

Answer 8

yes!

Answer 9

let u = cosx
du = -sinx dx
as sourwing said

Answer 10

yes Thats what I did

Answer 11

\[\int\limits_{ }^{ }-e^u du =-e^u =- e^\cosx \]

Answer 12

from 0 to pi/2

Answer 13

yeah

Answer 14

\[-e^cospi/2 +e^ \cos0\]

Answer 15

- isn't for all the equation?

Answer 16

ah oops sorry yeah

Answer 17

-e^0 +e^1=e-1

Answer 18

e= ?

Answer 19

omg I see now

Answer 20

^^
im a bit sleepy gn :D

Answer 21

My god , now, I dont know how I could thinks it was difficult hahaha :) good nught :)!!

Answer 22

na its not diffecult lol
by the way are u arabian ?

Answer 23

hahaha nooo! Canadian :)

Answer 24

But french :s

Answer 25

lol sry i told u im a bit sleepy XD
its 2:12 am So
Good night ^^

Answer 26

you yes?:o its 7h13 here !!