Verified answer

Ответ:

Применяем формулу синуса двойного угла  [tex]sin2\alpha =2\, sin\alpha \cdot cos\, \alpha[/tex]  и формулы понижения степеней:   [tex]sin^2\alpha =\dfrac{1-cos2\alpha }{2}[/tex]   , [tex]cos^2\alpha =\dfrac{1+cos2\alpha }{2}[/tex]  .  

[tex]\displaystyle \int\limits_ {-\frac{\pi}{2}}^{\frac{\pi }{2}}\, sin^2\frac{x}{4}\cdot cos^4\frac{x}{4}\, dx=\int\limits_ {-\frac{\pi}{2}}^{\frac{\pi }{2}}\Big(sin^2\frac{x}{4}\cdot cos^2\frac{x}{4}\Big)\cdot cos^2\frac{x}{4}\, dx=\\\\\\=\int\limits_ {-\frac{\pi}{2}}^{\frac{\pi }{2}}\, \frac{1}{4}\, sin^2\frac{x}{2}\cdot cos^2\frac{x}{4}\, dx=\frac{1}{4}\int\limits_ {-\frac{\pi}{2}}^{\frac{\pi }{2}}\, \frac{1-cosx}{2}\cdot \frac{1+cos\frac{x}{2}}{2}\, dx=[/tex]

[tex]\displaystyle =\frac{1}{16}\int\limits_ {-\frac{\pi}{2}}^{\frac{\pi }{2}}\, (1-cosx)\cdot (1+cos\frac{x}{2})\, dx=\frac{1}{16}\int\limits_ {-\frac{\pi}{2}}^{\frac{\pi }{2}}\, (1-cosx+cos\frac{x}{2}-cosx\cdot cos\frac{x}{2}\Big)\, dx=\\\\\\=\frac{1}{16}\cdot \Big(x-sinx+2sin\frac{x}{2}\Big)\Big|_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-\frac{1}{16}\int\limits_ {-\frac{\pi}{2}}^{\frac{\pi }{2}}\frac{1}{2}\cdot \Big(cos\frac{3x}{2}+cos\frac{x}{2}\Big)\, dx=[/tex]

[tex]\displaystyle =\frac{1}{16}\cdot \Big(x-sinx+2sin\frac{x}{2}\Big)\Big|_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-\frac{1}{8}\cdot \Big(\frac{2}{3}\, sin\frac{3x}{2}+2sin\frac{x}{2}\Big)\Big|_{-\frac{\pi }{2}}^{\frac{\pi}{2}}=\\\\\\=\frac{1}{16}\cdot \Big(\frac{\pi}{2}-1+2\cdot \frac{\sqrt2}{2}\Big)-\frac{1}{16}\cdot \Big(-\frac{\pi}{2} +1-2\cdot \frac{\sqrt2}{2}\Big)-[/tex]

[tex]\displaystyle -\frac{1}{8}\cdot \Big(\frac{2}{3}\cdot \frac{\sqrt2}{2}+2\cdot \frac{\sqrt2}{2}\Big)+\frac{1}{8}\cdot \Big(-\frac{2}{3}\cdot \frac{\sqrt2}{2}-2\cdot \frac{\sqrt2}{2}\Big)=\\\\\\=\frac{1}{8}\cdot \Big(\frac{\pi }{2}-1+\sqrt2\Big)-\frac{1}{4}\cdot \Big(\frac{\sqrt2}{3}+\sqrt2\Big)=\frac{1}{16}\cdot (\pi -2+2\sqrt2)-\frac{\sqrt2}{12}[/tex]