∫ \frac{cos2x.sin4x}{\cos^4x (1 + \cos^2 2x)} dx =

Correct Answer is : log(1+cos2x)2(1+cos22x)+sec2x+c

log (\frac{1+cos2x}{1+\cos^2 2x}) + \sec^2 x + c

log (1+cos2x)^2/(1+\cos^2 x) + sec x + c

\log {(1+\cos 2 x)^{2}}/{ (1+\cos ^{2} 2 x )}+ \sec ^{2} x+c

log (1+\cos^2 2x)/(1+\cos^2 2x)^2 + sec x + c

AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 74 of 160

Marks: +1, -0

∫

cos2x.sin4x

cos4x(1+cos22x)

dx =

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