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Optics
int e ^ (Sin^2 t ) dt ..................i.e. integral of e raised to Sin square t with respect to t
it shud b e^sin2t. cos2t/2
d above answer shud hv been wid a minus sign...
so d final answer is...
-e^sin2t. cos2t/2
Can you explain that thing???
I mean that I know that but still I think the answer is wrong.
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it shud b e^sin2t. cos2t/2