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

Question

25764 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-08T05:39:27

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

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 · 2018-06-08T06:49:03

Here is a Second Method to find the Integral.

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.!