Question

The value of the integral $\underset{0}{\overset{2\mathrm{\pi }}{\int }}{\cos }^{7}x{\sin }^{4}xdx$ is ________________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{0}{\overset{2\mathrm{\pi }}{\int }}{\cos }^{7}x{\sin }^{4}xdx \\ \mathrm{Let}f\left(x\right)={\cos }^{7}x{\sin }^{4}x \\ f\left(2\mathrm{\pi }-\mathrm{x}\right)={\left(\cos \left(2\mathrm{\pi }-\mathrm{x}\right)\right)}^7{\left(\sin \left(2\mathrm{\pi }-\mathrm{x}\right)\right)}^4 \\ ={\left(\cos \left(\mathrm{x}\right)\right)}^7{\left(-\sin \left(\mathrm{x}\right)\right)}^4 \\ ={\cos }^{7}x{\sin }^{4}x \\ =f\left(x\right) \\ \mathrm{We}\mathrm{know}, \\ \underset{0}{\overset{2a}{\int }}f\left(x\right)dx=\left\{\begin{array}{l}2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\mathrm{if}f\left(2a-x\right)=f\left(x\right)\\ 0,\mathrm{if}f\left(2a-x\right)=-f\left(x\right)\end{array}\right. \\ \mathrm{Thus}, \\ I=2\underset{0}{\overset{\mathrm{\pi }}{\int }}{\cos }^{7}x{\sin }^{4}xdx \\ \mathrm{Again}, \\ \mathrm{Let}f\left(x\right)={\cos }^{7}x{\sin }^{4}x \\ f\left(\mathrm{\pi }-\mathrm{x}\right)={\left(\cos \left(\mathrm{\pi }-\mathrm{x}\right)\right)}^7{\left(\sin \left(\mathrm{\pi }-\mathrm{x}\right)\right)}^4 \\ ={\left(-\cos \left(\mathrm{x}\right)\right)}^7{\left(\sin \left(\mathrm{x}\right)\right)}^4 \\ =-{\cos }^{7}x{\sin }^{4}x \\ =-f\left(x\right) \\ \mathrm{Thus}, \\ I=0 \end{gathered}$$​

​Hence, the value of the integral $\underset{0}{\overset{2\mathrm{\pi }}{\int }}{\cos }^{7}x{\sin }^{4}xdx$ is 0.