How do you integrate #intxsec^-1(x)dx#?

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Answer 1 · 2018-04-10T07:55:44

The answer is #=x^2/2arcsecx-1/2sqrt(x^2-1)+C#

#intuv'=uv-intu'v#

Here,

#u=arcsecx#, #=>#, #u'=(dx)/(xsqrt(x^2-1))#

#v'=x#, #=>#, #v=x^2/2#

Therefore,

#intxarcsecxdx=x^2/2arcsecx-1/2int(xdx)/(sqrt(x^2-1))#

Let #u=x^2-1#, #=>#, #du=2xdx#

Therefore,

#1/2int(xdx)/(sqrt(x^2-1))=1/4int(du)/(sqrtu)#

#=1/2sqrtu#

#=1/2sqrt(x^2-1)#

And finally,

#intxarcsecxdx=x^2/2arcsecx-1/2sqrt(x^2-1)+C#