作业帮1/(a*a+sinx*sinx)对x在[0,pi]的积分是多少可以用留数定理做！a是一个实常数！
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/03 07:31:09
![1/(a*a+sinx*sinx)对x在[0,pi]的积分是多少可以用留数定理做！a是一个实常数！](/uploads/image/z/10937343-39-3.jpg?t=1%2F%28a%2Aa%2Bsinx%2Asinx%29%E5%AF%B9x%E5%9C%A8%5B0%2Cpi%5D%E7%9A%84%E7%A7%AF%E5%88%86%E6%98%AF%E5%A4%9A%E5%B0%91%E5%8F%AF%E4%BB%A5%E7%94%A8%E7%95%99%E6%95%B0%E5%AE%9A%E7%90%86%E5%81%9A%EF%BC%81a%E6%98%AF%E4%B8%80%E4%B8%AA%E5%AE%9E%E5%B8%B8%E6%95%B0%EF%BC%81)
1/(a*a+sinx*sinx)对x在[0,pi]的积分是多少可以用留数定理做！a是一个实常数！
1/(a*a+sinx*sinx)对x在[0,pi]的积分是多少
可以用留数定理做！a是一个实常数！
1/(a*a+sinx*sinx)对x在[0,pi]的积分是多少可以用留数定理做！a是一个实常数！
我不是数学专业的,不懂留数定理.
下面用微积分基本公式提供一种思路,如有不对请不吝指教.
原理：∫dx/(a+bcosx)=2/(a+b)*√[(a+b)/(a-b)]*arctan[√[(a-b)/(a+b)]*tan(x/2)]+C (a*a>b*b)
原式=∫[1/(a*a+sinx*sinx)]dx (0,π) (√表示根号)
=∫[1/(a*a+(1-cos(2x))1/2)]dx(0,π)
=∫[1/(a*a+1/2-1/2cos(2x))]dx(0,π)
=1/2*∫[1/(c+dcos(y))]dy(0,π/2) (令c=a*a+1/2,d=-1/2,y=2x)
=1/2*[2/(c+d)*√[(c+d)/(c-d)]*arctan[√[(c-d)/(c+d)]*tan(y/2)] y∈(0,π/2)
=1/2*[2/(c+d)*√[(c+d)/(c-d)]*arctan[√[(c-d)/(c+d)])*tan(π/2/2)]
-1/2*[2/(c+b)*√[(c+d)/(c-d)]*arctan[√[(c-d)/(c+d)])*tan(0/2)]
=1/2*[2/(c+d)*√[(c+d)/(c-d)]*arctan[√[(c-d)/(c+d)]*1]
-1/2[2/(c+b)*√[(c+d)/(c-d)]*arctan[√[(c-d)/(c+d)]*0]
=1/2*[2/(c+d)*√[(c+d)/(c-d)]*arctan[√[(c-d)/(c+d)]*1]
还原c,d:
=1/2*[2/(a*a+1/2+（-1/2）)*√[(a*a+1/2+（-1/2）)/(a*a+1/2-(-1/2) ))]]*arctan[√[(a*a+1/2-(-1/2))/(a*a+1/2+(-1/2))]*1]
=1/2*[2/(a*a)*√[a*a/(a*a+1)]]*arctan√[(a*a+1)/(a*a)]
=1/(a*a)*√[(a*a/(a*a+1)]*arctan√[(a*a+1)/(a*a)]]
即为最终结果.