Evaluate the following integral. \( \int 9 \sin ^{-\frac{3}{2}} x \cos ^{3} x d x \) \( \int 9 \sin ^{-\frac{3}{2}} x \cos ^{3} x d x=\square \)

To evaluate the integral \( \int 9 \sin^{-\frac{3}{2}} x \cos^{3} x \, dx \), we can use a substitution approach along with integration techniques. First, let’s rewrite the integral: \[ I = \int 9 \sin^{-\frac{3}{2}} x \cos^{3} x \, dx \] We can use the substitution \( u = \sin x \), which gives us \( du = \cos x \, dx \). The integral transforms as follows: When \( u = \sin x \), then \( \cos x = \sqrt{1 - u^2} \). Therefore, \( \cos^3 x = (1 - u^2)^{3/2} \) and \( \sin^{-\frac{3}{2}} x = u^{-\frac{3}{2}} \). As a result, the differential \( dx = \frac{du}{\cos x} = \frac{du}{\sqrt{1 - u^2}} \). Substituting these into the integral, we have: \[ I = \int 9 u^{-\frac{3}{2}} (1 - u^2)^{\frac{3}{2}} \cdot \frac{du}{\sqrt{1 - u^2}} \] This simplifies to: \[ I = 9 \int u^{-\frac{3}{2}} (1 - u^2) \, du \] Now, we can distribute \( u^{-\frac{3}{2}} \): \[ I = 9 \left( \int u^{-\frac{3}{2}} \, du - \int u^{-\\frac{3}{2}} u^2 \, du \right) \] Calculating the first integral: \[ \int u^{-\frac{3}{2}} \, du = -2 u^{-\frac{1}{2}} + C_1 \] Now for the second integral: \[ \int u^{-\frac{3}{2}} u^2 \, du = \int u^{\frac{1}{2}} \, du = \frac{2}{3} u^{\frac{3}{2}} + C_2 \] Putting it all together: \[ I = 9 \left( -2 u^{-\frac{1}{2}} - \frac{2}{3} u^{\frac{3}{2}} \right) \] \[ = -18 u^{-\frac{1}{2}} - 6 u^{\frac{3}{2}} + C \] Now substituting back \( u = \sin x \): \[ I = -18 \sin^{-\frac{1}{2}} x - 6 \sin^{\frac{3}{2}} x + C \] Thus, the solution to the integral is: \[ \int 9 \sin^{-\frac{3}{2}} x \cos^{3} x \, dx = -18 \sin^{-\frac{1}{2}} x - 6 \sin^{\frac{3}{2}} x + C \]