1/(sinx ^3+cox ^3)的积分比较简单的做饭
时间: 2024-03-04 10:52:57 浏览: 160
这个积分可以用三角恒等式化简,得到:
1/(sin^3(x) + cos^3(x)) = 1/[(sin(x) + cos(x))(sin^2(x) - sin(x)cos(x) + cos^2(x))]
由于sin^2(x) + cos^2(x) = 1,因此:
sin^2(x) - sin(x)cos(x) + cos^2(x) = 1 - sin(x)cos(x)
将上面的式子代入分母中,得到:
1/(sin^3(x) + cos^3(x)) = 1/[(sin(x) + cos(x))(1 - sin(x)cos(x))]
我们可以将sin(x) + cos(x)表示成根号2的形式,即:
sin(x) + cos(x) = √2(sin(x + π/4))
令t = sin(x + π/4),则:
dt/dx = cos(x + π/4) = (cos(x) - sin(x))/√2
那么,原式可以变为:
∫1/[(sin(x) + cos(x))(1 - sin(x)cos(x))] dx = ∫1/[√2t(1 - t^2)] dt
对右侧的积分进行求解,可以使用换元法,令u = 1 - t^2,则:
du/dt = -2t,因此dt = -1/(2t) du
将u代入原式,得到:
∫1/[√2t(1 - t^2)] dt = -1/(2√2) ∫1/[tu^(1/2)] du
再次使用换元法,令v = u^(1/2),则:
dv/du = 1/(2u^(1/2)),因此du = 2v dv
将v代入原式,得到:
∫1/[tu^(1/2)] du = ∫1/(tv) 2v dv = 2∫1/t dt = 2ln|t| + C
将t代回,得到:
∫1/[(sin(x) + cos(x))(1 - sin(x)cos(x))] dx = -1/(√2) ln|1 - t^2| + C'
其中,C'为常数。将t代回,得到:
∫1/(sin^3(x) + cos^3(x)) dx = -1/(√2) ln|1 - sin^2(x + π/4)| + C'
这就是原式的解。
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