1. 原式=∫{1/√[ln(5x+1)}*[1/(5x+1)]*(1/5)*d(5x+1)
=∫{1/√]ln(5x+1)]}*(1/5)*d[ln(5x+1)]
=(1/5)*∫[ln(5x+1)]^(-1/2)*d[ln(5x+1)]
=(1/5)*[ln(5x+1)^(1/2)]/(1/2) + C
=2√[ln(5x+1)]/5 + C
2.设x=a*sint,则原积分的下限变为0,上限变为π/2
原式=∫(0,π/2) √(a^-(a*sint)^)*d(a*sint)
=∫(0,π/2) a√(1-sin^t)*a*d(sint)
=∫(0,π/2) a^cos^t dt
=∫(0,π/2) a^[(1+cos2t)/2] *dt
=∫(0,π/2) (a^/2 )*dt + ∫(0,π/2) (a^/2)*cos2t*dt
=(a^/2)*(π/2-0) + (a^/2)*[(sin2t)/2] (t下限为0,上限为π/2)
=a^π/4)
3.这个题为了表达清楚,我干脆先求不定积分,最后把上下限一带就可以!
原式=∫x*[(1-cos2x)/2]*dx
=∫xdx/2 - (1/2)*∫xcos2xdx
=x^/4 - (1/2)*∫(x/2)*d(sin2x)
=x^/4 - (1/4)*∫xd(sin2x)
=x^/4 - (1/4)*xsin2x + (1/4)*∫sin2xdx
=x^/4 - xsin2x/4 - cos2x/8
代入上下限0,π的值,可得到:
原式=π^/4 - 1/8 - (-1/8)
=π^/4
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