Question

Помогите решить интеграл

Answer 1

$sin^2x= \frac{1-cos2x}{2} \\ \int\limits^{ \frac{\pi}{4} }_0 {\frac{1-cos2x}{2} } \, dx = \left[\begin{array}{ccc}2x=u\\2dx=du\\dx= \frac{du}{2} \end{array}\right]= \int\limits^{ \frac{\pi}{4} }_0 { \frac{1-cosu}{4}}\, du= \frac{1}{4} \int\limits^{ \frac{\pi}{4} }_0{(1-cosu)}\,du = \\= \frac{1}{4}*(u-sinu) \int\limits^{ \frac{\pi}{4} }_0=\frac{1}{4}*(2x-sin2x) \int\limits^{ \frac{\pi}{4} }_0= \frac{1}{4} (\frac{\pi}{2}-sin \frac{\pi}{2}-0)= \\= \frac{1}{4} (\frac{\pi}{2}-1)= \frac{\pi-2}{8}$