## Calculus (3rd Edition)

$$\frac{1}{n+1}$$
Given $$\int_{0}^{\pi / 2} \sin ^{n} x \cos x d x$$ For $n=0$ $$\int_{0}^{\pi / 2} \sin ^{n} x \cos x d x=\int_{0}^{\pi / 2} \cos x d x=\sin x \bigg|_{0}^{\pi/2}=1$$ For n$\gt0$, let $$u=\sin x \ \ \ \Rightarrow \ \ \ du = \cos x dx$$ At $$x= 0 \to u= 0, \ \ x= \pi/2\to u= 1$$ Then \begin{aligned} \int_{0}^{\pi / 2} \sin ^{n} x \cos x d x &=\int_{0}^{1} u^{n} d u \\ &=\left.\frac{u^{n+1}}{n+1}\right|_{0} ^{1} \\ &=\frac{1}{n+1} \end{aligned}