tích phân ( sin x/2 + cos x/2 )^2 dx bằng:

14/50

\[\int {{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}{\rm{d}}x} \] bằng:

\[x - \cos x + C\].

\[{\left( { - \cos \frac{x}{2} + \sin \frac{x}{2}} \right)^2} + C\].

\[\frac{1}{3}{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)^3} + C\].

\[x + \cos x + C\].

Giải thích

Ta có: \[{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)^2} = {\sin ^2}\frac{x}{2} + 2\sin \frac{x}{2} \cdot \cos \frac{x}{2} + {\cos ^2}\frac{x}{2} = 1 + \sin x\].

Khi đó \[\int {{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}{\rm{d}}x} = \int {\left( {1 + \sin x} \right){\rm{d}}x} = \int {{\rm{d}}x} + \int {\sin x{\rm{d}}x = x - \cos x + C} \]. Chọn A.