Question

How do you evaluate sqrt(64 + x^2)/x?

Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)

`int` `sqrt(64 + x^2)/x` dx

Answer 1

`I = sqrt(x^2 +64) - 8ln|sqrt(x^2 + 64)/x + 8/x| + C`

Explanation:

Use the trig substitution `x = 8tantheta`. This means that `dx = 8sec^2theta d theta`

`I = int sqrt(64 + (8tantheta)^2)/(8tantheta) * 8sec^2theta d theta`

`I = int sqrt(64 + 64tan^2theta)/(8tantheta) * 8sec^2theta d theta`

`I = int sqrt(64(1 + tan^2theta))/(8tantheta) * 8sec^2theta d theta`

`I = int sqrt(64sec^2theta)/tantheta sec^2theta d theta`

`I = int (8sec^3theta)/tantheta d theta`

`I = int 8sec^3theta cot theta d theta`

`I = 8 int 1/costheta(sec^2theta)costheta/sintheta d theta`

`I = 8 int sec^2theta/sintheta d theta`

`I = 8int sec^2theta csctheta d theta`

`I = 8 int (1 + tan^2theta)csctheta d theta`

`I = 8int csctheta + sin^2theta/cos^2theta(1/sintheta) d theta`

`I = 8 int cscthetad theta + 8int sintheta/cos^2theta d theta`

`I = 8int csctheta d theta + 8 inttanthetasectheta d theta`

These are two known integrals. We know that `int cscx dx = -ln(cscx + cotx)` and `int secxtanxdx = secx`. Thus:

`I = 8sectheta - 8ln|csctheta + cottheta| + C`

From our initial substitution, we know that `tantheta = x/8`. Therefore, if we were to draw a right triangle we would find the hypotenuse to be `sqrt(x^2 + 8^2) = sqrt(x^2 + 64)`. Thus, `sectheta = sqrt(x^2 + 64)/8`, `csctheta = sqrt(x^2+ 64)/x` and `cottheta = 8/x`.

`I = 8sqrt(x^2 + 64)/8 - 8ln|sqrt(x^2+ 64)/x + 8/x| + C`

`I = sqrt(x^2 +64) - 8ln|sqrt(x^2 + 64)/x + 8/x| + C`

Hopefully this helps!