How do you integrate #Sinx * Tanx#?

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Answer 1 · 2016-12-09T15:08:01

The answer is #=ln (∣tanx+secx∣)-sinx +C#

We need, #secx=1/cosx#

#cos^2x+sin^2x=1#

#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#