If $I = \int e^{x} s i n 2 x d x$ , then for what value of K, $K I = e^{x} (s i n 2 x - 2 c o s 2 x) + c o n s t a n t$

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Solution

The correct option is C 5
$I = \int e^{x} s i n 2 x d x = s i n 2 x . e^{x} - 2 \int c o s 2 x . e^{x} d x = s i n 2 x . e^{x} - 2 c o s 2 x . e^{x} - 4 \int e^{x} s i n 2 x d x \Rightarrow 5 I = e^{x} (s i n 2 x - 2 c o s 2 x) + c o n s t a n t$
Equating the given value, we get K = 5

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