tìm tích phân sin bình x/2 dx
Giải thích
\(\int {{{\sin }^2}} \frac{x}{2}\;{\rm{d}}x = \int {\frac{{1 - \cos x}}{2}} \;{\rm{d}}x = \frac{1}{2}\left( {\int {\rm{d}} x - \int {\cos } x\;{\rm{d}}x} \right) = \frac{1}{2}(x - \sin x) + C\)
\(\int {{{\sin }^2}} \frac{x}{2}\;{\rm{d}}x = \int {\frac{{1 - \cos x}}{2}} \;{\rm{d}}x = \frac{1}{2}\left( {\int {\rm{d}} x - \int {\cos } x\;{\rm{d}}x} \right) = \frac{1}{2}(x - \sin x) + C\)