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Integral Calculus

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9 Nov 2008 20:53:32 IST

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integrate under root cos2x divided by sinx

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integrate under root cos2x divided by sinx

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Himanshu

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Joined: 19 Feb 2007

Posts: 3834

10 Nov 2008 01:03:38 IST

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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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