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Discussion Response Post to: integrate under root cos2x divided by sinx

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Himanshu (9616)

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= integ (1 - 2 sin²x / sinx)

seperating these 2 we get

I = (dx / (sinx(cos²x - sin²x)^1/2) - 2 (sinx / (2cos²x - 1)^1/2)

= (cosec²x / (cot²x - 1)^1/2) - (2/root2) (sinx/(cos²x - (1/2)^1/2))

take cot x = a cos x = b and solve u will get the answer....

(p.s-> formula editor not working of slow speed...soo...manage...plzz).

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