evaluate the integral.\n int_{pi}^{3pi/2} 45sin^{2}xcos^{3}x dx \n int_{pi}^{3pi/2} 45sin^{2}xcos^{3}x…

Explanation:

Step1: Rewrite $\cos^{3}x$

We know that $\cos^{3}x=\cos x\cos^{2}x=\cos x(1 - \sin^{2}x)$. So the integral $\int_{\pi}^{\frac{3\pi}{2}}45\sin^{2}x\cos^{3}x dx=45\int_{\pi}^{\frac{3\pi}{2}}\sin^{2}x\cos x(1 - \sin^{2}x)dx$.

Step2: Use substitution

Let $u = \sin x$, then $du=\cos xdx$. When $x=\pi$, $u=\sin\pi = 0$; when $x=\frac{3\pi}{2}$, $u=\sin\frac{3\pi}{2}=- 1$. The integral becomes $45\int_{0}^{-1}u^{2}(1 - u^{2})du=45\int_{0}^{-1}(u^{2}-u^{4})du$.

Step3: Integrate term - by - term

Using the power rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)$, we have $45\left[\frac{u^{3}}{3}-\frac{u^{5}}{5}\right]_{0}^{-1}$.

Step4: Evaluate the definite integral

First, substitute $u=-1$ into $\frac{u^{3}}{3}-\frac{u^{5}}{5}$: $\frac{(-1)^{3}}{3}-\frac{(-1)^{5}}{5}=-\frac{1}{3}+\frac{1}{5}=\frac{-5 + 3}{15}=-\frac{2}{15}$. Then substitute $u = 0$ (which gives $0$). So $45\times\left(-\frac{2}{15}\right)=-6$.

Answer:

$-6$