What is dx of sin x/2cos^2x.
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Posted by Mayank Shahabadee 9 years, 2 months ago
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Rashmi Bajpayee 9 years, 2 months ago
dx = sin x / 2cos2x = 1/2[sin x - sec2x] [By Product rule]
dx = 1/2[sin x.2sec x.sec x.tan x + sec2x.cos x]
dx = 1/2[sin x.2sec2x.tan x + sec x]
dx = (1/2) sec x[2sin2x.sec2x + 1] = (1/2) sec x[2tan2x + 1] = 1/(2cos x) [(2sin2x/cos2x) + 1]
dx = 1/(2cos3 x) [2sin2x + 1 - sin2x ] = 1/(2cos3 x) [sin2x + 1]
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