# Using trigonometric substitution, find `int(sqrt(9x^2-25))/(x^3)dx` where `x>5/3`I know that for `sqrt(x^2-a^2)`  it's x=a*sec`Theta` . But I have no idea where to go from there.

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to factor out 9 under the square root, such that:

`int sqrt(9(x^2 - (5/3)^2))/x^3 dx`

You need to come up with the substitution such that:

`x = (5/3)/(cos t) => x^2 = (5/3)^2/(cos t)^2`

`dx = (5/3 sin t)/(cos t)^2 dt`

Changing the variable yields:

`3 int sqrt((5/3)^2/(cos t)^2 - (5/3)^2)/((5/3)^3/(cos t)^3) (5/3 sin t)/(cos t)^2 dt`

Factoring out `(5/3)^2` yields:

`3*(5/3) int sqrt(1 - cos^2 t)/(cos t*(5/3)^2)*(sin t* cos t) dt`

Using the fundamental formula of trigonometry `1 - cos^2 t = sin^2 t` and reducing duplicate factors yields:

`9/5 int sin^2 t dt`

You need to use the following trigonometric identity such that:

`sin^2 t = (1 - cos 2t)/2`

`9/5 int (1 - cos 2t)/2dt`

Using the linearity of integral yields:

`9/5 int (1 - cos 2t)/2dt = 9/5 int (1/2) dt - 9/5 int (cos 2t)/2dt`

`9/5 int (1 - cos 2t)/2dt = (9/10) t - (9/20) sin 2t + c`

Substituting back `cos^(-1)(3x/5)` for `t` yields:

`int sqrt(9(x^2 - (5/3)^2))/x^3 dx = (9/10)cos^(-1)(3x/5) - (9/20) sin (2cos^(-1)(3x/5)) + c`

Hence, evaluating the given indefinite integral, using the trigonometric substitution, yields `int sqrt(9(x^2 - (5/3)^2))/x^3 dx = (9/10)cos^(-1)(3x/5) - (9/20) sin (2cos^(-1)(3x/5)) + c.`