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hi I've some question regarding cal.3
1- how can i calculate the double integral of xcos(2x+y)dA
0<= x <= 2pi/6 , 0<=y<=pi/4
that is what i have done:
xcos(2x+y)dy = xsin(2x+y) = xsin(2x+pi/4) - xsin(2x)
then how i can integrate dx?
your help is very much appericated
First, a not on what was given: I'm unclear how x•sin(2x+y) is equal to
x•sin(2x+pi/4) - x•sin(2x). According to an addition of angles formula in trig,
sin(2x+y) = sin(2x)cos(y) + sin(y)cos(2x).
But to solve the double integral, I assume the dA is really dx•dy.
To integrate with respect to x, use the u-v rule and let u = x, so du = 1,
and let dv = cos(2x+y)dx, so v = sin(2x+y)/2.
To integrate with respect to y, ∫∫x•cos(2x+y) dx dy is x•sin(2x+y) since
the derivative of sin(2x+y), by the chain rule, with respect to y, is cos(2x+y).
After doing these integrals, put in the upper limit and
subtract off what is gotten by putting in the lower limit.
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