Question

$$\underset{0}{\overset{\pi }{\int }}x\sin x{\cos }^{4}xdx$$

Answer 1

$$Let,I={\int }_0^{\pi }x\sin x{\cos }^{4}xdx............(1)$$

$$={\int }_0^{\pi }(\pi -x)\sin (\pi -x){\cos }^4(\pi -x)dx$$

$$={\int }_0^{\pi }(\pi -x)\sin x{\cos }^{4}xdx..............(2)$$

Adding (1) and (2)

$$2I={\int }_0^{\pi }[x\sin x{\cos }^{4}x+(\pi -x)\sin x{\cos }^{4}x]dx$$

$$={\int }_0^{\pi }(x+\pi -x)\sin x{\cos }^{4}xdx$$

$$=\pi {\int }_0^{\pi }\sin x{\cos }^{4}xdx$$

$$=\pi {\left[\frac{-{\cos }^{5}x}{5}\right]}_0^{\pi }$$

$$=\pi [\frac{1}{5}+\frac{1}{5}]$$

$$=\frac{2\pi }{5}$$

$$Hence,I=\frac{\pi }{5}$$