Answer to Question #189876 in Calculus for Moel Tariburu

Question

Evaluate the integral ∫▒〖sin〗^3 x 〖cos〗^4 xdx 

Answer 1

$\int sin^3xcos^4xdx=\int -cos^4x(cos^2x-1)sinxdx$

u=cosx

du/dx=-sinxdx=-du/sinx

$\int -cos^4x(cos^2x-1)sinxdx=\int u^4(u^2-1)du=\int (u^6-u^4)du=\int u^6du-\int u^4du=u^7/7-u^5/5=cos^7x/7-cos^5x/5+C$

Answer:$\int sin^3xcos^4xdx=cos^7x/7-cos^5x/5+C$