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RolyPoly
 2 years ago
Best ResponseYou've already chosen the best response.0\[sin\omega t * cos \omega t\]\[=\int_0^tsin\omega u \ cos \omega (tu)du\]\[=\frac{1}{\omega}\int_0^t\ cos \omega (tu)d(cos\omega u)\]\[=\frac{1}{2\omega}[ cos ^2\omega (tu)]_0^t\] Doesn't seem right

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1\[ =\int_0^t \sin\omega u \; \cos \omega (tu)du \\ =\int_0^t \sin\omega u \; (\cos \omega t \cos \omega u  \sin \omega t \sin \omega u)du \\ = \int_0^t \sin\omega u \; \cos \omega t \;\cos \omega u \;du  \int_0^t \sin \omega u \;\sin \omega t \; \sin \omega u \;du \]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1\[ =\int_0^t \sin\omega u \; \cos \omega (tu)du \\ =\int_0^t \sin\omega u \; (\cos \omega t \cos \omega u  \sin \omega t \sin \omega u)du \\ = \cos \omega t \; \int_0^t \sin\omega u \;\cos \omega u \;du  \sin \omega t \;\int_0^t \sin \omega u \; \sin \omega u \;du \]
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