How do you integrate $sin x \cdot tan x$ $Sinx * Tanx$ ?

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Answer 1 · 2016-12-09T15:08:01

The answer is $= ln (∣ tan x + sec x ∣) - sin x + C$ $=ln (∣tanx+secx∣)-sinx +C$

We need, $sec x = \frac{1}{cos x}$ $secx=1/cosx$

${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x+sin^2x=1$

$tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$

$(tanx)'=sec^2x$

$(secx)'=tanx secx$

$intsinxtanxdx=int(sinx*sinxdx)/cosx$

$=intsecxsin^2xdx$

$=intsecx(1-cos^2x)dx$

$=int(secx-cosx)dx=intsecxdx-intcosxdx$

For the integral of $secx$ , multiply top and bottom by
$(tanx+secx)$

$intsecxdx=int(secx(tanx+secx)dx)/(tanx +secx)$

Let $u=tanx +secx$

$du=(sec^2x+secxtanx)dx$

$intsecxdx=int(du)/u$

$=lnu = ln (∣tanx+secx∣)$

And finally,

$intsinxtanxdx= ln (∣tanx+secx∣)-sinx +C$

How do you integrate Sinx * Tanxsinx⋅tanxSinx * Tanx?