x*(sin(x))的平方的积分
时间: 2024-05-17 07:17:31 浏览: 159
以sinx为例的近似矩形公式求定积分
这个积分可以通过分部积分法来求解。
设u=sin(x),dv= x*sin(x)dx,则du=cos(x)dx,v=-x*cos(x)+sin(x)
则原积分可以表示为:
∫x*(sin(x))^2dx = -x*cos(x)*(sin(x))^2 + ∫cos(x)*(sin(x))^2dx
再次应用分部积分法,设u=sin(x),dv=sin(x)cos(x)dx,则du=cos^2(x)dx,v=-cos(x)cos(x)/2
则原积分可以进一步化简为:
∫x*(sin(x))^2dx = -x*cos(x)*(sin(x))^2 - cos(x)*(sin(x))^2/2 + ∫cos^2(x)*(sin(x))^2dx
化简最后一项,利用三角恒等式sin^2(x)=1-cos^2(x),得到:
∫cos^2(x)*(sin(x))^2dx = ∫(1-sin^2(x))*(sin(x))^2cos(x)dx
令u=sin(x),dv=sin(x)cos(x)dx,则du=cos^2(x)dx,v=-cos(x)cos(x)/2
则∫(1-sin^2(x))*(sin(x))^2cos(x)dx = ∫(1-u^2)u^2du = ∫(u^2-u^4)du = u^3/3 - u^5/5
将上述结果代入原积分中,得到:
∫x*(sin(x))^2dx = -x*cos(x)*(sin(x))^2 - cos(x)*(sin(x))^2/2 + u^3/3 - u^5/5
最终答案为:
-x*cos(x)*(sin(x))^2 - cos(x)*(sin(x))^2/2 + (sin(x))^3/3 - (sin(x))^5/5 + C,其中C为积分常数。
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