Question

How do you integrate `int sqrt(x^2-25)/x dx` using trigonometric substitution?

Answer 1

`sqrt(x^2-25)-5arcsec(x/5)+C.`

Explanation:

Suppose that, `I=intcolor(red)sqrt(x^2-25)/color(green)xdx`, and, let, `color(green)(x=5secu)`.

`:. color(blue)(dx=5secutanudu)`.

`:. I=intcolor(red)sqrt(25sec^2u-25)/color(green)(5secu)*color(blue)(5secutanudu)`,

`=intcolor(red)(5tanu)*color(blue)(tanudu)`,

`=5inttan^2udu`,

`=int(sec^2u-1)du`,

`=5(tanu-u)`.

Here, `color(red)(5tanu=sqrt(x^2-25),) and, color(green)(x=5secu rArr u=arcsec(x/5).`

`rArr I=sqrt(x^2-25)-5arcsec(x/5)+C,` as desired!

Enjoy Maths.!

Answer 2

Here is a Second Method to find the Integral.

Explanation:

Let, `I=intsqrt(25-x^2)/xdx=intsqrt{(x^2-25)/x^2}dx`.

`:. I=intsqrt(1-(5/x)^2)dx`.

We subst. `5/x=sint. :. sqrt(1-(5/x)^2)=cost`.

`"Also, "5/x=sint. :. x=5csct. :. dx=-5csctcottdt`.

`:. I=int(cost)(-5csctcott)dt`,

`=-5int{cost*1/sint*cost/sint}dt`,

`=-5intcot^2tdt`,

`=-5int(csc^2t-1)dt`,

`=-5(-cott-t)`,

`=5{(cost/sint)+t}`,

`=5{sqrt(1-(5/x)^2)/(5/x)+arcsin(5/x)}`.

`rArr I=sqrt(x^2-25)+5arcsin(5/x)+C`.

I leave it to the Questioner to show that both the Answers are

equivalent!

Enjoy Maths.!