Question

求三角函数多次方积分∫（0到2π）sin(t/2)的5次方dt,怎么求积分

Answer 1

要求∫_0^2pi▒〖(sin⁡(t/2)⁡ )^5〗dt,

先化sin(t/2)的四次方,(sin(t/2))^4=(1-(cos(t/2))^2)^2=1-2(cos(t/2))^2+(cos(t/2))^4,令t/2=x,所以x在(0,pi),先化不定积分2∫(sin(x))^5dx=-2∫(1-cos(x)^2)^2d(cos(x))=-2∫（1-2(cos(x))^2+(cos(x))^4）d(cos(x))=-2(cos(x)-2/3cos(x)^3+1/5cos(x)^5)+C所以∫_0^2pi▒〖(sin⁡(t/2)⁡ )^5〗dt=28/15.

只简单的算了下 结果不一定正确,我认为重在方法是怎么做就可以啦!