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1. I = ∫dx/[(sinx)^2(cosx)^2] = ∫2d2x/(sin2x)^2 = -2cot2x + C
2. I = ∫sinxdx/(sinx+cosx) = ∫sinxdx/[√2sin(x+π/4)] ,令 u = x+π/4,
I = (1/√2)∫sin(u-π/4)du/sinu = (1/2)∫(1-cosu/sinu)du
= (1/2)(u-lnsinu) + C1 = (1/2)(x-lnsin(x+π/4) + C
3. I = (1/20)∫(5x^4-1)^(1/2)d(5x^4-1) = (1/30)(5x^4-1)^(3/2) + C
4. I = (-1/2)∫sin(1/x^2)d(1/x^2) = (1/2)cos(1/x^2) + C
5. 令 u = √x, 则 I = ∫(u^2+1)2du = (2/3)u^3 + 2u + C = 2√x(x/3 +1) + C
二. f(x) = 2x-2
I = ∫xf(1-x^2)dx = ∫x[2(1-x^2)-2]dx = -2∫x^3dx = (-1/2)x^4 + C
2. I = ∫sinxdx/(sinx+cosx) = ∫sinxdx/[√2sin(x+π/4)] ,令 u = x+π/4,
I = (1/√2)∫sin(u-π/4)du/sinu = (1/2)∫(1-cosu/sinu)du
= (1/2)(u-lnsinu) + C1 = (1/2)(x-lnsin(x+π/4) + C
3. I = (1/20)∫(5x^4-1)^(1/2)d(5x^4-1) = (1/30)(5x^4-1)^(3/2) + C
4. I = (-1/2)∫sin(1/x^2)d(1/x^2) = (1/2)cos(1/x^2) + C
5. 令 u = √x, 则 I = ∫(u^2+1)2du = (2/3)u^3 + 2u + C = 2√x(x/3 +1) + C
二. f(x) = 2x-2
I = ∫xf(1-x^2)dx = ∫x[2(1-x^2)-2]dx = -2∫x^3dx = (-1/2)x^4 + C
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