![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
如何求解∫(sinxcos
1个回答
展开全部
∫ (sinxcosx)/(sinx + cosx) dx=(1/2)(- cosx + sinx) - [1/(2√2)]ln|csc(x + π/4) - cot(x + π/4)| + C。C为积分常数。
解答过程如下:
∫ (sinxcosx)/(sinx + cosx) dx
= (1/2)∫ (2sinxcosx)/(sinx + cosx) dx
= (1/2)∫ [(1 + 2sinxcosx) - 1]/(sinx + cosx) dx
= (1/2)∫ (sin²x + 2sinxcosx + cos²x)/(sinx + cosx) dx - (1/2)∫ dx/(sinx + cosx)
= (1/2)∫ (sinx + cosx)²/(sinx + cosx) dx - (1/2)∫ dx/[√2sin(x + π/4)]
= (1/2)∫ (sinx + cosx) dx - [1/(2√2)]∫ csc(x + π/4) dx
= (1/2)(- cosx + sinx) - [1/(2√2)]ln|csc(x + π/4) - cot(x + π/4)| + C
解答过程如下:
∫ (sinxcosx)/(sinx + cosx) dx
= (1/2)∫ (2sinxcosx)/(sinx + cosx) dx
= (1/2)∫ [(1 + 2sinxcosx) - 1]/(sinx + cosx) dx
= (1/2)∫ (sin²x + 2sinxcosx + cos²x)/(sinx + cosx) dx - (1/2)∫ dx/(sinx + cosx)
= (1/2)∫ (sinx + cosx)²/(sinx + cosx) dx - (1/2)∫ dx/[√2sin(x + π/4)]
= (1/2)∫ (sinx + cosx) dx - [1/(2√2)]∫ csc(x + π/4) dx
= (1/2)(- cosx + sinx) - [1/(2√2)]ln|csc(x + π/4) - cot(x + π/4)| + C
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
![](https://ecmc.bdimg.com/public03/b4cb859ca634443212c22993b0c87088.png)
2024-11-18 广告
广东聚氨酯板材认准佛山市泰升塑胶科技有限公司,专业聚氨酯包胶厂家,获国家证书认证,按客户标准,欢迎来样定做.具有高回弹,高耐磨,耐酸碱,抗水解等特性,规格多,价格低,服务好。 佛山市泰升塑胶科技有限公司是研制、生产聚氨酯塑胶系列产品的专业企...
点击进入详情页
本回答由佛山泰升塑胶公司科技提供
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