How do you integrate #int sqrt(-x^2-6x-18)/xdx# using trigonometric substitution?

Question

How do you integrate #int  sqrt(-x^2-6x-18)/xdx# using trigonometric substitution?

Emily Ammerman · Accepted Answer

Complete the square under the #sqrt#. #int sqrt(-(x^2 + 6x + 9 - 9) - 18)/xdx# #int sqrt(-(x^2 + 6x + 9) + 9 - 18)/xdx# #int sqrt(-(x + 3)^2 - 9)/xdx# Let #u = x + 3#.  Then #du = dx#. #int sqrt(-u^2 -9)/(u - 3) du# But the expression under the #sqrt# clearly has no real values of #x# that satisfy.  So techniqually this problem has no solution. Hopefully this helps!

Ezra Adams · Answer

undefined To integrate $\int \frac{\sqrt{-x^2-6x-18}}{x} \, dx$ using trigonometric substitution, we first complete the square in the denominator. This yields $\int \frac{\sqrt{-x^2-6x-18}}{x} \, dx = \int \frac{\sqrt{-x^2-6x-18}}{x} \cdot \frac{x}{x} \, dx = \int \frac{\sqrt{x^2+6x+18}}{x} \, dx$. Then, we let $x = -3 \sec(\theta)$ so that $dx = -3 \sec(\theta) \tan(\theta) \, d\theta$. Substituting these expressions into the integral and simplifying, we obtain $\int \frac{\sqrt{x^2+6x+18}}{x} \, dx = \int \frac{\sqrt{9 \sec^2(\theta)}}{-3 \sec(\theta)} \cdot (-3 \sec(\theta) \tan(\theta)) \, d\theta = \int -3 \tan^2(\theta) \, d\theta$. Integrating this expression with respect to $\theta$, we find $\int -3 \tan^2(\theta) \, d\theta = -3 \int \tan^2(\theta) \, d\theta$. Using the identity $\tan^2(\theta) = \sec^2(\theta) - 1$, we rewrite the integral as $-3 \int (\sec^2(\theta) - 1) \, d\theta$. Integrating term by term, we obtain $-3 \left(\tan(\theta) - \theta \right) + C$, where $C$ is the constant of integration. Finally, we substitute back $x = -3 \sec(\theta)$ and simplify to find the final result of the integral.