Evaluate the integral. \( \int_{3 \pi / 2}^{2 \pi} 45 \sin ^{2} x \cos ^{3} x d x \) \( 2 \pi / 2 \sin ^{2} x \cos ^{3} x d x=6 \) \( ( \) Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) \( 2 \pi \)

To evaluate the integral \( \int_{3 \pi / 2}^{2 \pi} 45 \sin ^{2} x \cos ^{3} x \, dx \), we can use the substitution method. Notice that if we let \( u = \cos x \), then \( du = -\sin x \, dx \), which transforms the integral involving \( \sin^2 x \) and \( \cos^3 x \). First, determine the new limits of integration: when \( x = 3\pi/2 \), \( u = \cos(3\pi/2) = 0 \), and when \( x = 2\pi \), \( u = \cos(2\pi) = 1 \). The integral becomes: \[ \int_{0}^{1} 45 (1 - u^2)(u^3)(-du) = 45 \int_{0}^{1} (u^3 - u^5) \, du \] Now we can integrate: \[ 45 \left[ \frac{u^4}{4} - \frac{u^6}{6} \right]_{0}^{1} = 45 \left( \frac{1}{4} - \frac{1}{6} \right) = 45 \left( \frac{3}{12} - \frac{2}{12} \right) = 45 \cdot \frac{1}{12} = \frac{45}{12} = \frac{15}{4} \] Thus, the value of the integral is: \[ \frac{15\pi}{4} \] This means the correct answer, simplified (as integers or fractions), is \( \frac{45}{12} \) or \( 3.75 \) if that's what you prefer! Now go grab a slice of pie – it's not every day you get to play with pi in integrals! Wouldn't it be fun to apply this technique while calculating areas of wacky-shaped pies? It's integral to your pastry endeavors!