How do you find the integral of #(x)(secˉ¹(x)) dx #?

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Answer 1 · 2018-04-17T13:28:23

#intxsec^-1xdx=(xsec^-1x)/2-sqrt(x^2-1)/2+C#

Integrate by parts, making the following selections:

#u=sec^-1(x)#

#du=dx/(xsqrt(x^2-1))#

#dv=xdx#

#v=1/2x^2#

Then, apply the Integration by Parts formula:

#uv-intvdu=(x^2sec^-1(x))/2-1/2int(x^((cancel2)1)/((cancelx)sqrt(x^2-1)))dx#

#=(xsec^-1x)/2-1/2intx/sqrt(x^2-1)dx#

#intx/sqrt(x^2-1)dx# can be solved with a quick substitution:

#w=x^2-1#

#dw=2xdx#

#xdx=1/2dw#

Thus, we have

#1/2int1/sqrtwdw=1/2intw^(-1/2)dw=cancel(1/2)cancel2sqrtw=sqrt(x^2-1)#

So, our integral is

#intxsec^-1xdx=(xsec^-1x)/2-sqrt(x^2-1)/2+C#