大一高数定积分 求详细过程 50
3个回答
展开全部
1-cost=2[sin(t/2)]^2
(1-cost)^(5/2)=4√2*[sin(t/2)]^5
原式=8a^3*∫(0,2π)[sin(t/2)]^5dt
令x=t/2,则t=2x,dt=2dx
原式=16a^3*∫(0,π)(sinx)^5dx
=-16a^3*∫(0,π)(1-cos^2x)^2d(cosx)
=-16a^3*∫(0,π)(1-2cos^2x+cos^4x)d(cosx)
=-16a^3*[x-(2/3)*cos^3x+(1/5)*cos^5x]|(0,π)
=-16a^3*(π+2/3-1/5+2/3-1/5)
=-16*(π+14/15)a^3
(1-cost)^(5/2)=4√2*[sin(t/2)]^5
原式=8a^3*∫(0,2π)[sin(t/2)]^5dt
令x=t/2,则t=2x,dt=2dx
原式=16a^3*∫(0,π)(sinx)^5dx
=-16a^3*∫(0,π)(1-cos^2x)^2d(cosx)
=-16a^3*∫(0,π)(1-2cos^2x+cos^4x)d(cosx)
=-16a^3*[x-(2/3)*cos^3x+(1/5)*cos^5x]|(0,π)
=-16a^3*(π+2/3-1/5+2/3-1/5)
=-16*(π+14/15)a^3
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