计算定积分,要求有必要具体过程,题目内容见图.
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计算定积分,要求有必要具体过程,题目内容见图.
![](http://img.wesiedu.com/upload/9/fb/9fbbbb94b9566e4cef7241d4055f04d2.jpg)
![](http://img.wesiedu.com/upload/9/fb/9fbbbb94b9566e4cef7241d4055f04d2.jpg)
![计算定积分,要求有必要具体过程,题目内容见图.](/uploads/image/z/18425450-2-0.jpg?t=%E8%AE%A1%E7%AE%97%E5%AE%9A%E7%A7%AF%E5%88%86%2C%E8%A6%81%E6%B1%82%E6%9C%89%E5%BF%85%E8%A6%81%E5%85%B7%E4%BD%93%E8%BF%87%E7%A8%8B%2C%E9%A2%98%E7%9B%AE%E5%86%85%E5%AE%B9%E8%A7%81%E5%9B%BE.)
x = tanu,dx = sec²u du
x = 1,u = π/4
x = √3,u = π/3
[1,√3] ∫ dx / [x² √(1+x²)]
= [π/4,π/3] ∫ sec²u du / [tan²u √(1 + tan²u)]
= [π/4,π/3] ∫ cosu du / sin²u
= [π/4,π/3] ∫ dsinu / sin²u
= -1/sinu | [π/4,π/3]
=1/sin(π/4) - 1/sin(π/3)
=√2 - 2√3 /3
x = 1,u = π/4
x = √3,u = π/3
[1,√3] ∫ dx / [x² √(1+x²)]
= [π/4,π/3] ∫ sec²u du / [tan²u √(1 + tan²u)]
= [π/4,π/3] ∫ cosu du / sin²u
= [π/4,π/3] ∫ dsinu / sin²u
= -1/sinu | [π/4,π/3]
=1/sin(π/4) - 1/sin(π/3)
=√2 - 2√3 /3