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Question
Evaluate:
`int(cos 2x)/sinx dx`
Sum
Solution
Let I = `int(cos 2x)/sinx dx`
∴ I = `int(1 - 2sin^2x)/sinx dx`
= `int(1/sinx - (2sin^2x)/sinx)dx`
= `int("cosec" x - 2sinx)dx`
= log (cosec x – cot x) + 2 cos x + c
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