Evaluate the following integral ∫〖〖cos〗^3 x〗 〖sin〗^4 x dx

∫〖〖cos〗^3 x〗  〖sin〗^4 x dx=∫〖〖cos〗^2 x〗  〖sin〗^4 x cos⁡x  dx
=∫(1-〖sin〗^2 x)  〖sin〗^4 x cos⁡x  dx
If we let =sin⁡x   ,then du=cos⁡x  dx , and the integral may be written:
∫〖〖cos〗^3 x〗  〖sin〗^4 x dx=∫〖(1-u^2 ) u^4 〗 du=∫(u^4-u^6 )  du
=1/5 u^5-1/7 u^7+c
=1/5 〖sin〗^5 x-1/7 〖sin〗^7 x+c