高数大一
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/08/19 15:34:20
高数大一
![](http://img.wesiedu.com/upload/d/73/d73f307ca2ea12271c0fc0660cd75694.jpg)
![](http://img.wesiedu.com/upload/d/73/d73f307ca2ea12271c0fc0660cd75694.jpg)
![高数大一](/uploads/image/z/6858121-49-1.jpg?t=%E9%AB%98%E6%95%B0%E5%A4%A7%E4%B8%80)
答:
1)
∫ [2/(x-3)]dx
=∫ [2/(x-3)]d(x-3)
=2ln|x-3|+C
2)
(0→π/2) ∫ (cosx)^2dx
=(0→π/2) ∫ (cos2x+1)/2 dx
=(0→π/2) (1/4)*∫(cos2x+1)d(2x)
=(0→π/2) (sin2x+2x)/4
=(0+π)/4-0
=π/4
3)
∫ (tanx)^2 dx
=∫ [(sinx)^2/(cosx)^2] dx
=∫ {[1-(cosx)^2]/(cosx)^2} dx
=∫ [1/(cosx)^2-1]dx
=∫ [(secx)^2-1]dx
=tanx-x+C
此题应用到下面的公式:
![](http://img.wesiedu.com/upload/c/21/c2130ca271d0a5782c9c0752d81c4a96.jpg)
1)
∫ [2/(x-3)]dx
=∫ [2/(x-3)]d(x-3)
=2ln|x-3|+C
2)
(0→π/2) ∫ (cosx)^2dx
=(0→π/2) ∫ (cos2x+1)/2 dx
=(0→π/2) (1/4)*∫(cos2x+1)d(2x)
=(0→π/2) (sin2x+2x)/4
=(0+π)/4-0
=π/4
3)
∫ (tanx)^2 dx
=∫ [(sinx)^2/(cosx)^2] dx
=∫ {[1-(cosx)^2]/(cosx)^2} dx
=∫ [1/(cosx)^2-1]dx
=∫ [(secx)^2-1]dx
=tanx-x+C
此题应用到下面的公式:
![](http://img.wesiedu.com/upload/c/21/c2130ca271d0a5782c9c0752d81c4a96.jpg)