Evaluate the following integral ∫〖〖cos〗^3 x〗 〖sin〗^4 x dx
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Mathematics
∫〖〖cos〗^3 x〗 〖sin〗^4 x dx=∫〖〖cos〗^2 x〗 〖sin〗^4 x cosx dx
need an explanation for this answer? contact us directly to get an explanation for this answer=∫(1-〖sin〗^2 x) 〖sin〗^4 x cosx dx
If we let =sinx ,then du=cosx 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