Question #2250e

1 Answer
Mar 13, 2017

1/2"arcsec"^2(x)+C

Explanation:

I=int("arcsec"(x))/(xsqrt(x^2-1))dx

Let x=sectheta. This implies that dx=secthetatanthetad theta. Also it means that theta="arcsec"(x). Then:

I=inttheta/(secthetasqrt(sec^2theta-1))(secthetatanthetad theta)

Note that sec^2theta-1=tan^2theta:

I=inttheta/(secthetatantheta)(secthetatantheta)d theta

I=intthetacolor(white).d theta

I=1/2theta^2+C

I=1/2"arcsec"^2(x)+C