How do you find the integral of #tanh^3x dx#?

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Answer 1 · 2018-08-12T15:29:29

#inttanh(x)^3dx=ln(|cosh(x)|)+1/2sech(x)^2+C#, #C in RR#

#I=inttanh(x)^3dx#

#=intsinh(x)^3/(cosh(x)^3)dx#

Because #sinh(x)^2=cosh(x)^2-1#,

#I=int(sinh(x)(cosh(x)^2-1))/(cosh(x)^3)dx#

Now let #u=cosh(x)#

#du=sinh(x)dx#

So:

#I=int(u^2-1)/u^3du#

#=int1/udu-int1/u^3du#

#=ln(|u|)+1/(2u^2)+C#, #C in RR#

#=ln(|cosh(x)|)+1/2sech(x)^2+C#, #C in RR#

\0/ Here's our answer !