How do you evaluate the integral: #sqrt(x^2-a^2)/x dx#?

For the integral involving the root #sqrt(x^2-a^2),# we use the substitution:

#x=asectheta#

#dx=asecthetatanthetad theta#

So, we get

#int(cancelacancelsecthetatanthetasqrt(a^2(sec^2theta-1)))/(cancelacancelsectheta))d theta#

Recalling the identity #sec^2theta-1=tan^2theta#, we get

#intatanthetasqrt(tan^2theta)=intatan^2thetad theta#

To integrate this, we'll use the identity #tan^2theta=sec^2theta-1# again.

#inta(sec^2theta-1)d theta=intasec^2thetad theta-intad theta=atantheta-atheta+C#

We need to get things in terms of #x.# Recalling that #x=asectheta, sectheta=x/a, theta=sec^-1(x/a)#

To find the tangent, we'll use the identity

#tan^2theta=sec^2theta-1#

#tan^2theta=x^2/a^2-1#

#tantheta=sqrt(x^2-a^2)/a#

Thus,

#intsqrt(x^2-a^2)/xdx=(asqrt(x^2-a^2))/a-asec^-1(x/a)+C#

#intsqrt(x^2-a^2)/xdx=sqrt(x^2-a^2)-asec^-1(x/a)+C#

Here,

#I=intsqrt(x^2-a^2)/xdx#

Let ,#x=asecu=>dx=asecutanudu#

#and secu=x/a=>u=sec^-1(x/a)#

#:.I=intsqrt(a^2sec^2u-a^2)/cancel((asecu))(cancel(asecu)tanu)du#

#=intsqrt(a^2tan^2u)*tanudu#

#=int(atanu)*tanudu#

#=ainttan^2udu#

#=aint(sec^2u-1)du#

#=a(tanu-u)+c#

#=a(sqrt(sec^2u-1)-u)+c#, where, #secu=x/a,u=sec^-1(x/a)#

#=a[sqrt((x/a)^2-1)-sec^-1(x/a)]+c#

#=a[sqrt(x^2-a^2)/a-sec^-1(x/a)]+c#

#=sqrt(x^2-a^2)-asec^-1(x/a)+c#