Question

double integral of cos xy dydx from 0 to 1/x and 1 to 4

Answer 1

\[\int_1^4\int_0^{1/x}\cos(xy)~dy~dx\]
Substitute \(u=xy\), then \(\dfrac{du}{dy}=x\), so that \(\dfrac{1}{x}du=dy\):
\[\int_1^4\int_0^1\frac{1}{x}\cos u~du~dx\]

Answer 2

It becomes \[\int\limits_{1}^{4} -sinx dx\] Right?

Answer 3

No, integrating with respect to \(u\) first yields
\[\int_1^4\frac{1}{x}\bigg[\sin u\bigg]_0^1~dx=\int_1^4\frac{1}{x}(\sin 1-0)~dx=\sin1\int_1^4\frac{dx}{x}\]

Answer 4

Makes sense now. Was kind of confused at first but I'm catching on. Thanks a lot.

Answer 5

You're welcome!