Trig Substitution D36C1F

1. **State the problem:** Rewrite the integral $$\int \frac{\sqrt{16x^2 - 81}}{x^3} \, dx$$ using the substitution $$x = \frac{9}{4} \sin \theta$$. 2. **Recall the substitution and its derivative:** Given $$x = \frac{9}{4} \sin \theta$$, then $$dx = \frac{9}{4} \cos \theta \, d\theta$$. 3. **Rewrite the expression inside the square root:** $$16x^2 - 81 = 16 \left(\frac{9}{4} \sin \theta\right)^2 - 81 = 16 \cdot \frac{81}{16} \sin^2 \theta - 81 = 81 \sin^2 \theta - 81 = 81 (\sin^2 \theta - 1) = -81 \cos^2 \theta$$. 4. **Simplify the square root:** $$\sqrt{16x^2 - 81} = \sqrt{-81 \cos^2 \theta} = 9i \cos \theta$$. Since the integral involves a real variable, we consider the absolute value and the domain of $$\theta$$ such that $$\cos \theta$$ is positive or negative accordingly. For the substitution, we take $$\sqrt{16x^2 - 81} = 9 |\cos \theta|$$. 5. **Rewrite the denominator:** $$x^3 = \left(\frac{9}{4} \sin \theta\right)^3 = \frac{729}{64} \sin^3 \theta$$. 6. **Substitute all parts into the integral:** $$\int \frac{\sqrt{16x^2 - 81}}{x^3} \, dx = \int \frac{9 |\cos \theta|}{\frac{729}{64} \sin^3 \theta} \cdot \frac{9}{4} \cos \theta \, d\theta = \int \frac{9 \cdot 9}{\frac{729}{64} \sin^3 \theta} |\cos \theta| \cos \theta \cdot \frac{1}{4} \, d\theta$$. 7. **Simplify constants:** $$\frac{9 \cdot 9 \cdot 64}{729 \cdot 4} = \frac{81 \cdot 64}{2916} = \frac{5184}{2916} = \frac{64}{36} = \frac{16}{9}$$. 8. **Simplify the integral:** $$\int \frac{16}{9} \frac{|\cos \theta| \cos \theta}{\sin^3 \theta} \, d\theta$$. Assuming $$\cos \theta \geq 0$$ in the domain, $$|\cos \theta| = \cos \theta$$, so $$\int \frac{16}{9} \frac{\cos^2 \theta}{\sin^3 \theta} \, d\theta$$. **Final rewritten integral:** $$\int \frac{16}{9} \frac{\cos^2 \theta}{\sin^3 \theta} \, d\theta$$. This is the exact expression for the integral after substitution.