
高数题, 求不定积分题 希望可以写在纸上,写出详细的步骤,我在线等,
展开全部
sinx = sin[2*(x/2)] = 2sin(x/2)cos(x/2)
1 = cos^2(x/2) + sin^2(x/2)
1 + sinx = cos^2(x/2) + 2sin(x/2)cos(x/2) + sin^2(x/2) = [cos(x/2) + sin(x/2)]^2
后面就很简单了
1 = cos^2(x/2) + sin^2(x/2)
1 + sinx = cos^2(x/2) + 2sin(x/2)cos(x/2) + sin^2(x/2) = [cos(x/2) + sin(x/2)]^2
后面就很简单了
追问
嗯,看懂了
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询