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Решите задачу:

$\displaystyle\int{ \sin^{4}(\frac{x}{2})}dx =\begin{vmatrix}\frac{x}{2}=u\\du=\frac{dx}{2}\end{vmatrix}=2\int{{\sin(u)}^{4}du}=\frac{1}{2}\int{{(1-\cos(2u))}^{2}du}=\frac{1}{2}\int{1-2\cos(2u) +\cos(2u)^{2}du} =\frac{1}{2}(u-\sin(2u)+\int{\frac{1+\cos(4u)}{2}du}=\frac{1}{2}(u+\frac{1}{4}\sin(2u)+\frac{1}{2}(u+\frac{\sin(4u)}{4})+C=\frac{u}{2}-\frac{1}{2}\sin(2u)+\frac{u}{4}+\frac{\sin(4u)}{16}+c=\frac{x}{2}+\frac{x}{8}-\frac{1}{2}\sin(x)+\frac{\sin(2x)}{8}+C=\frac{3x}{8}-\frac{1}{2}\sin(x)+\frac{1}{16}\sin(2x)+C$