求 Sin 2021X 比上 sin x 在 0~2 分之派上的定积分。
创建于:2025年2月17日
创建于:2025年2月17日
求 Sin 2021X 比上 sin x 在 0~2 分之派上的定积分。
Interpreting the query
The user seems to want to compute the definite integral of sin(x)sin(2021x) from 0 to 22π, but there's some uncertainty in the translation. "求" means "find," and "sin 2021X 比上 sin x" translates as sin(x)sin(2021x). The “0~2 分之派” likely means "from 0 to π," as "2分之π" is a common notation for 2π. So, the calculation should be from 0 to π.
Clarifying the integral range
The user’s question involves sin(x)sin(2021x), with a limit of integration from 0 to possibly 2π (since "2分之派" translates to 2π) or maybe 2π. There’s a known formula that ∫0πsin(x)sin(mx)dx=π when m is odd, and the result is zero if m is even. Based on this, if the interval is from 0 to π, the result would be π. If it's from 0 to 2π, the result would be 2π.
Evaluating the integral limits
The user is trying to evaluate an integral of sinxsin(2021x), and there’s some confusion about the interval. The phrase "2分之派" could mean 2π, so the interval might be from 0 to 2π. There's an identity for ∫0πsin(x)sin((2n+1)x)dx=π, and adjusting this for our specific case, it seems the integral from 0 to 2π will give 2π.
Reasoned for 19 seconds
我们要求解积分
I=∫0π/2sinxsin(2021x)dx.注意到 2021 是奇数,我们可以写成
2021=2n+1其中n=1010.利用恒等式
sinxsin((2n+1)x)=1+2k=1∑ncos(2kx),积分便转化为
I=∫0π/2[1+2k=1∑1010cos(2kx)]dx.对 1 的积分很简单:
∫0π/21dx=2π.而对于每个 k≥1,有
∫0π/2cos(2kx)dx=2ksin(2kx)0π/2=2ksin(kπ)=0,因为 sin(kπ)=0。
因此,所有余弦项的积分均为零,从而
I=2π+2k=1∑10100=2π.故答案为 2π.