Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 276: 81

$$\frac{1}{4}$$

Given $$ \int_{0}^{\pi / 2} \cos ^{3} x \sin x d x$$ Let $$ u= \cos x \ \ \ \Rightarrow \ \ \ du =-\sin xdx$$ At $$ x= 0\to u= 1, \ \ x= \pi/2\to u= 0$$ Then \begin{aligned} \int_{0}^{\pi / 2} \cos ^{3} x \sin x d x&=-\int_{1}^{0} u^{3} d u\\ &=\int_{0}^{1} u^{3} d u\\ &= \frac{1}{4}u^4\bigg|_{0}^{1}\\ &= \frac{1}{4} \end{aligned}