How do you evaluate int dx/(x^2sqrt(x^2 - 9)) with x = 3sec(theta)?

Evaluate the integral using the indicated trigonometric substitution. (Use C for the constant of integration.)

int dx/(x^2sqrt(x^2 - 9)), x = 3sec(theta)

1 Answer
Jul 5, 2018

I=int dx/(x^2sqrt(x^2 - 9))

Let x = 3sectheta

=>dx = 3sectheta tantheta d theta

So
I=int(3sectheta tantheta d theta)/(9sec^2thetasqrt(9sec^2theta- 9))

=int(3sectheta tantheta d theta)/(27sec^2thetasqrt(sec^2theta- 1))

=1/9int(sectheta tantheta d theta)/(sec^2theta tan theta)

=1/9intcosthetad theta

=1/9sintheta +C

=1/9tantheta/sectheta +C

=1/9(sqrt(sec^2theta-1))/sectheta +C

=1/9(sqrt(x^2/9-1))/(x/3) +C

=1/9(sqrt(x^2-9))/x +C