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

1 Answer
Apr 17, 2018

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

Explanation:

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