Question

∫∫sin (xy+y) dA. where R= {(x,y)|0<=x<=pi, 0<=y<=pi/2 how to do the integration?

Answer 1

is there a trig indentity for sin (a+b)?

Answer 2

just an idea, dunno how useful it may be tho

Answer 3

x and y are in cartesian coordinates? Just to be sure

Answer 4

yes.

Answer 5

sin (xy+y) = sin(xy)cos(y)+sin(y)cos(xy)

this creates a constant in y so that you only have to deal with the x parts first

Answer 6

but.....still confused any simple example

Answer 7

\[\int_{0}^{pi/2}\int_{0}^{pi}sin (xy+y)~ dxdy\]\[\int_{0}^{pi/2}\int_{0}^{pi}sin(xy)cos(y)+sin(y)cos(xy)~ dxdy \]

since y is a constant with respect to x ....

\[\int_{0}^{pi/2}\int_{0}^{pi}sin(Cx)A+Bcos(Cx)~ dxdy \]

Answer 8

\[\int_{0}^{pi/2}\int_{0}^{pi}sin(xy)cos(y)+sin(y)cos(xy)~ dxdy\]
\[\int_{0}^{pi/2}\frac1ycos(\frac{pi}2 y)cos(y)+\frac1ysin(y)sin(\frac{pi}2y)-\frac1ycos(y)~dy\]
might have kept track of all that

Answer 9

is there any other method? method by substution ?

Answer 10

there are always other methods .....

Answer 11

ok. if possible able to let me knw..

Answer 12

system froze; but i see a few errors; sin comes from -cos, and i used the wrong limits ....

Answer 13

ok....

Answer 14

so, focusing on the innards ....
\[\int_{0}^{pi}sin(xy)cos(y)+sin(y)cos(xy)~ dx\]
\[\left.-\frac1ycos(xy)cos(y)+\frac1ysin(y)sin(xy)\right|_{0}^{pi}\]
\[-\frac1ycos(pi~y)cos(y)+\cancel{\frac1ysin(y)sin(pi~y)}^0+\frac1y\cancel{cos(0y)}^1 cos(y)-\cancel{\frac1ysin(y)sin(0y)}^0\]
\[\frac1ycos(y)(1-cos(pi~y))\]
hmmm

Answer 15

maybe easier to rewrite it at this stage

\[-\frac1y(cos(xy)cos(y)-sin(y)sin(xy))\]
\[-\left.\frac1ycos(xy+y)\right|_{0}^{pi}\]

Answer 16

\[−\frac1ycos((1+pi)~y)+\frac1ycos(y))\]
\[\int_{0}^{pi/2}−\frac1ycos((1+pi)~y)+\frac1ycos(y))~dy\]

Answer 17

ok.

Answer 18

http://www.wolframalpha.com/input/?i=integrate%28integrate+sin%28xy%2By%29%2C+x%3D0..pi%29%2C+y%3D0..pi%2F2

ewww, that aint pretty no matter what

Answer 19

i cant get a nice run on it ....

Answer 20

its ok. thanks lot for the steps.

Answer 21

good luck with it ;)

Answer 22

also, make sure youve typed the question in correctly