How do you integrate #sin(x)tan(x)#?

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Answer 1 · 2018-05-25T16:30:28

#(sin(x))^3/3+C#

Writing your integral in the form #\int sin(x)^2/cos(x)dx# and substitute#t=sin(x)#
then we get
#dx=dt/cos(x)#

Answer 2 · 2018-05-25T16:55:53

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

The integral is

#I=intsinxtanxdx=int(sin^2xdx)/cosx#

#=int(1-cos^2x)secxdx#

#=int(secx-cosx)dx#

#=intsecxdx-intcosxdx#

#=I_1+I_2#

First calculate

#I_1=intsecxdx#

#=int(secx(secx+tanx)dx)/(secx+tanx)#

#=int((sec^2x+secxtanx)dx)/(secx+tanx)#

Let #u=secx+tanx#, #=>#, #du=(secxtanx+sec^2x)dx#

Therefore,

#I_1=int(du)/u=lnu#

#=ln(secx+tanx)#

Second calculate

#I_2=intcosxdx=sinx#

And finally,

#I=ln(|secx+tanx|)-sinx+C#