Question

CAN ANYONEEEEEEEEEEE
integrate : Xlogsinx dx
from 0 to pi

Answer 1

wolfram can :)

http://www.wolframalpha.com/input/?i=integrate+xlog%28sin%28x%29%29+dx+from+0+to+pi

Answer 2

no
it doesnot show step by step

Answer 3

hmm indefinite integral can't be done using elementary functions

Answer 4

use the property \(\large \int \limits_a^bf(x)dx=\int \limits_a^bf(a+b-x)dx\)
and see how much it simplifies!

Answer 5

ofcourse you have to add the 2 equations.....

Answer 6

I =int Xlogsinx dx --------->(1)
I =int (pi- X)logsinx dx ----->(2)
ADD (1) and (2)
what u get ?

Answer 7

pi logsinx

Answer 8

2i=pilogsinx dx

Answer 9

yes,
now you can refer this :
http://openstudy.com/users/aliza_k#/updates/5039a6e2e4b043c156a31ff9

Answer 10

?? im lost
\[\int\limits_{0}^{\pi} x \ln(\sin x) dx \ne \int\limits_{0}^{\pi} \pi \ln(\sin x) dx\]

Answer 11

its 2I = int pi*log sin

Answer 12

oh ok

Answer 13

doubts ?
i have used that property ...and then added....ask if you still didn't get it...

Answer 14

no doubt
thank u so much @hartnn

Answer 15

welcome ^_^

Answer 16

omg it's too lengthy

Answer 17

yes, it is.
What you have posted is not an easy integral to solve.
But after using that property, it reduces to classic definite integral problem (ln sin x), which has a classic solution i explained there.

Answer 18

k

Answer 19

my answer is cmg
-pi/2 (log2)

Answer 20

there,
I =I/2 -pi/2 ln 2
so, I/2 = pi/2 ln 2
I = pi ln 2

Answer 21

-pi^2/2 (log2)

Answer 22

oh, wait....
yes thats correct
-pi^2/2 (log2) = -3.4205
which wolf gave....