# Evaluate the integral integrate of (dx)/(x^2(sqrt(x^2-36)))

### 1 Answer | Add Yours

You should use the following substitution such that:

`x = 6/cos t => dx = 6sin t/cos t dt = 6 tan t dt `

Changing the variable at integrand yields:

`int 1/(x^2sqrt(x^2-36))dx = int (cos^2 t)/(36sint sqrt(36/(cos^2 t) - 36)) dt`

Factoring out 36 yields:

`int (cos^2 t)/(36 sint sqrt(36/(cos^2 t) - 36)) dt = int (cos^2 t)/(36 sint sqrt(36(1/(cos^2 t) - 1))) dt`

You should use the following trigonometric identity such that:

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

`int (cos^2 t)/(216 sint sqrt(tan^2 t)) dt = 1/(216) int (cos^2 t)/tan t dt`

`1/(216) int(cos^2 t)/tan t dt = 1/(216) int (cos^2 t)/tan t dt`

`1/(216) int (cos^2 t)/tan t dt = 1/(216) int (cos^3 t)/sin t dt`

Using `1 - sin^2 t = cos^2 t ` yields:

`1/(216) int (1 - sin^2 t)cos t/sin t dt`

Using the suubstitution `sin t = u => cos t dt = du`

`1/(216) int (1 - sin^2 t)cos t/sin t dt = 1/(216) int (1 - u^2)/u du`

Using the linearity yields:

`1/(216) int (1 - u^2)/u du = 1/(216) (int (1/u)du - int (u^2)/u du)`

`1/(216) int (1 - u^2)/u du = 1/216 ln|u| - u^2/432 + c`

Substituting back `sin t` for `u` yields:

`1/(216) int (1 - sin^2 t)cos t/sin t dt = 1/216 ln|sin t| - (sin^2 t)/432 + c`

Substituting back `(sec^(-1) x)/6` for t yields:

`int 1/(x^2sqrt(x^2-36))dx = 1/216 ln|sin((sec^(-1) x)/6)| - (sin^2 ((sec^(-1) x)/6))/432 + c`

**Hence, evaluating the given integral, using trigonometric identities yields `int 1/(x^2sqrt(x^2-36))dx = 1/216 ln|sin((sec^(-1) x)/6)| - (sin^2 ((sec^(-1) x)/6))/432 + c.` **