How do you integrate $y = \frac{{sin}^{2} x}{cos x}$ $y= sin^2x/cosx$ ?

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Answer 1 · 2016-05-02T13:54:10

$ln (sec x + tan x) - sin x + C$ $ln (sec x + tan x) - sin x + C$

$y = \frac{1 - {cos}^{2} x}{cos x} = sec x - cos x$ $y=(1-cos^2x)/cos x= secx - cos x$

$= (ln( sec x + tan x ))'-(sin x)'$ .

So, $int y dx=ln( sec x + tan x) -sin x+C$ .

How do you integrate y= sin^2x/cosxy=sin2xcosxy= sin^2x/cosx?