Trig Substitution A0D8C4

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. **Find $dx$ in terms of $d\theta$: ** Given $$x = \frac{9}{4} \sin \theta,$$ then $$dx = \frac{9}{4} \cos \theta \, d\theta$$. 3. **Rewrite the expression under 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 is real, consider the absolute value and domain of $\theta$ so that \[ \sqrt{16x^2 - 81} = 9 |\cos \theta| \] 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:** \[ \int \frac{\sqrt{16x^2 - 81}}{x^3} \, dx = \int \frac{9 |\cos \theta|}{\frac{729}{64} \sin^3 \theta} \cdot \frac{9}{4} \cos \theta \, d\theta \] 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:** Assuming $\cos \theta \geq 0$ in the domain, so $|\cos \theta| = \cos \theta$, then \[ \int \frac{16}{9} \frac{\cos^2 \theta}{\sin^3 \theta} \, d\theta \] **Final rewritten integral:** $$\int \frac{16}{9} \frac{\cos^2 \theta}{\sin^3 \theta} \, d\theta$$