求下列定积分∫(上限pai/3i,下限0)(x/cos^2) dx;∫(pai/2,0) (x+sinx)/(1+cos

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  • ∫(0~π/3) x/cos²x dx = ∫(0~π/3) xsec²x dx

    = ∫(0~π/3) x d(tanx)

    = xtanx - ∫(0~π/3) tanx dx

    = (π/3)tan(π/3) - (- lncosx)

    = (π/3)(√3) + lncos(π/3) - lncos(0)

    = π/√3 + ln(1/2) - ln(1)

    = π/√3 - ln(2)

    ∫(0~π/2) (x + sinx)/(1 + cosx) dx

    = ∫(0~π/2) x/(1 + cosx) dx + ∫(0~π/2) sinx/(1 + cosx) dx

    = ∫(0~π/2) x/(1 + cosx) dx + ∫(0~π/2) tan(x/2) dx

    = ∫(0~π/2) x/(1 + cosx) dx + xtan(x/2) - ∫(0~π/2) x d(tan(x/2))

    = ∫(0~π/2) x/(1 + cosx) dx + (π/2)tan(π/4) - ∫(0~π/2) x/(1 + cosx) dx

    = π/2

    ∫(0~2) 1/[2 + √(4 - x²)] dx,令x = 2sinz,dx = 2cosz dz

    = ∫(0~π/2) (2cosz)/(2 + 2cosz) dz

    = ∫(0~π/2) cosz/(1 + cosz) dz

    = ∫(0~π/2) cosz/(1 + 2cos²(z/2) - 1) dz

    = (1/2)∫(0~π/2) [cos²(z/2) - sin²(z/2)]/cos²(z/2) dz

    = (1/2)∫(0~π/2) [1 - tan²(z/2)] dz

    = (1/2)∫(0~π/2) dz - (1/2)∫(0~π/2) [sec²(z/2) - 1] dz

    = (1/2)(π/2) - (1/2)∫(0~π/2) sec²(z/2) dz + (1/2)∫(0~π/2) dz

    = π/4 - tan(z/2) + π/4

    = π/2 - [tan(π/4) - tan(0)]

    = π/2 - 1

    = (π - 2)/2

    ∫(0~π/2) √(1 - sin2x) dx

    = ∫(0~π/2) √(sin²x - 2sinxcosx + cos²x) dx

    = ∫(0~π/2) |sinx - cosx| dx

    = ∫(0~π/4) (cosx - sinx) dx + ∫(π/4~π/2) (sinx - cosx) dx

    = (sinx + cosx) + (- cosx - sinx)

    = {[sin(π/4) + cos(π/4)] - [sin(π/2) + cos(π/2)]} + {[- cos(π/2) - sin(π/2)] - [- cos(π/4) - sin(π/4)]}

    = 2(√2 - 1)

    ∫(0~π/2) 1/(1 + sin²x) dx

    = ∫(0~π/2) 1/(sin²x + cos²x + sin²x) dx

    = ∫(0~π/2) 1/(cos²x + 2sin²x) dx

    = ∫(0~π/2) sec²x/(1 + 2tan²x) dx,上下除以cos²x

    = ∫(0~π/2) 1/(1 + 2tan²x) d(tanx)

    = (1/√2)∫(0~π/2) 1/(1 + (√2tanx)²) d(√2tanx)

    = (1/√2)arctan(√2tanx)

    = (1/√2)arctan[√2tan(π/2)] - (1/√2)arctan[√2tan(0)]

    = (1/√2)(π/2) - 0

    = π/(2√2)