Question

求下列定积分

∫（上限pai/3i,下限0）(x/cos^2) dx;
∫(pai/2，0) (x+sinx)/(1+cosx) dx;
∫(2,0) 1/[2+(4-x^2)^1/2] dx;
∫(pai/2，0) (1-sin2x)^1/2 dx;
∫(pai/2，0) 1/(1+sinx^2) dx

Answer 1

∫(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)