Question

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

Answer 1

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

Explanation:

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`

Answer 2

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

Explanation:

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`