判断定积分的正负,如图…
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/07/04 16:00:09
判断定积分的正负,如图…
![](http://img.wesiedu.com/upload/8/2d/82dc32d829911811c0f073e3f4dfd045.jpg)
![](http://img.wesiedu.com/upload/8/2d/82dc32d829911811c0f073e3f4dfd045.jpg)
![判断定积分的正负,如图…](/uploads/image/z/4795218-18-8.jpg?t=%E5%88%A4%E6%96%AD%E5%AE%9A%E7%A7%AF%E5%88%86%E7%9A%84%E6%AD%A3%E8%B4%9F%2C%E5%A6%82%E5%9B%BE%E2%80%A6)
【0,π】∫(1+sin²t)d(sin2t)=【0,π】[sin2t+∫sin²td(sin2t)]=【0,π】∫sin²td(sin2t)
=【0,π】(1/2)∫(1+cos2t)d(sin2t)=【0,π】[(1/2)sin2t+(1/2)∫(cos2td(sin2t)]
=【0,π】(1/2)∫(cos2td(sin2t)=【0,π】(1/2)[cos2tsin2t+2∫sin²2tdt]
=【0,π】∫sin²2tdt=【0,π】(1/2)∫(1+cos4t)dt=【0,π】(1/2)[t+(1/4)∫cos4td(4t)]
=(1/2)[t+(1/4)sin4t]【0,π】=(1/2)π
是正值.
=【0,π】(1/2)∫(1+cos2t)d(sin2t)=【0,π】[(1/2)sin2t+(1/2)∫(cos2td(sin2t)]
=【0,π】(1/2)∫(cos2td(sin2t)=【0,π】(1/2)[cos2tsin2t+2∫sin²2tdt]
=【0,π】∫sin²2tdt=【0,π】(1/2)∫(1+cos4t)dt=【0,π】(1/2)[t+(1/4)∫cos4td(4t)]
=(1/2)[t+(1/4)sin4t]【0,π】=(1/2)π
是正值.