Tích phân từ 0đến π của ( sin( x)^ 8)

$\int\limits^\pi_0 {sin^8x} \, dx$

=$\int\limits^\pi_0 {(\frac{1-cos2x}{2})^4} \, dx$

=$\int\limits^\pi_0 {\frac{1}{16}(1-4cos2x+6cos^22x-4cos^32x+cos^42x)} \, dx$

=$\int\limits^\pi_0 {\frac{1}{16}(1-4cos2x+6cos^22x-4cos^32x+(\frac{1+cos2x}{2})^2} )\, dx$

=$\int\limits^\pi_0 {\frac{1}{16}(1-4cos2x+6cos^22x-4cos^32x+\frac{1}{4}} +\frac{cos2x}{2}+\frac{cos^22x}{4})\, dx$ 

=$\int\limits^\pi_0 {\frac{1}{16}(\frac{5}{4}-\frac{7}{2}cos2x+\frac{25}{4}cos^22x-4cos^32x} )\, dx$

=$\int\limits^\pi_0 {\frac{1}{16}(\frac{5}{4}-\frac{7}{2}cos2x+\frac{25(cos4x+1)}{8}-4(1-sin^22x)cos2x} )\, dx$  

=$\int\limits^\pi_0 {\frac{1}{16}(\frac{35}{8}-\frac{15}{2}cos2x+\frac{25cos4x}{8}+4sin^22xcos2x} )\, dx$ 

=$\frac{1}{16}(\frac{35x}{8}-\frac{15sin2x}{4}+\frac{25sin4x}{32}+\frac{2sin^32x}{3})|^\pi_0$

=$\frac{35\pi}{128}$