Answer
$\int_1^2 \int_0^{1/x} \cos (xy) \ dy \ dx=\sin (1) \ln (2)$
Work Step by Step
Here, we have: $\int_1^2 \int_0^{1/x} \cos (xy) \ dy \ dx=\int_1^2[\dfrac{\sin (xy)}{x}]_0^{1/x} \ dx$
or, $= \int_1^2 [\dfrac{\sin (1)}{x}] \ dx$
or, $=\sin (1) \times [\ln (x)]_1^2$
or, $=\sin (1) \times [\ln (2)-\ln 1]$
or, $=\sin (1) \times [\ln (2)-0]$
Thus, we get:
$\int_1^2 \int_0^{1/x} \cos (xy) \ dy \ dx=\sin (1) \ln (2)$
