Question

这道题怎么做，中间的

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Answer 1

令y = sech^-1(x),则x = sech(y),dx = [sech(y)]'dy
而d[sech^-1(x)]/dx = dy/dx = 1/[sech(y)]'
sech(y) = 2/(e^y + e^-y)
[sech(y)]' = [2/(e^y + e^-y)]' = -2(e^y - e^-y)/(e^y + e^-y)² = -sinh(y)/cosh²(y)
dy/dx = 1/[sech(y)]' = -cosh²(y)/sinh(y)
因为x = sech(y),所以cosh(y) = 1/x,sinh(y) = √(1 - 1/x²)
代入上式得到dy/dx = -1/x² * 1/√(1 - 1/x²) = -1/[x√(x²-1)]
即d[sech^-1(x)]/dx = -1/[x√(x²-1)]
4.
∫(-3,3)e^xt dt
= 1/x * ∫(-3,3)e^xt dxt
= 1/x * e^xt|(-3,3)
= 1/x * (e^3x - e^-3x)
= 2sinh(3x)/x