Question

求∫(0-π/2)e^2xcosxdx，∫(0-π)(xsinx)^2dx，∫(π/3-π/4)x/sinx^2dx，∫(0-π^2/4)sin(x^1/2)定积分

Answer 1

解：
∫(0-π/2)e^(2x)cosxdx＝∫(0-π/2)e^(2x)dsinx＝e^(2x)sinx|(0-π/2)－∫(0-π/2)sinxde^(2x)＝e^π·sin(π/2)－0－2∫(0-π/2)e^(2x)sinxdx＝e^π＋2∫(0-π/2)e^(2x)dcosx＝e^π＋2[e^(2x)cosx|(0-π/2)－∫(0-π/2)cosxde^(2x)]＝e^π＋2e^π·cos(π/2)－2e^0·cos0－4∫(0-π/2)e^(2x)cosxdx＝e^π－2－4∫(0-π/2)e^(2x)cosxdx
取连等式最左端和最右端：∫(0-π/2)e^(2x)cosxdx＝e^π－2－4∫(0-π/2)e^(2x)cosxdx
∴5∫(0-π/2)e^(2x)cosxdx＝e^π－2
∴∫(0-π/2)e^(2x)cosxdx＝(e^π－2)/5

∫(0-π)(xsinx)²dx＝1/2∫(0-π)x²·2sin²xdx＝1/2∫(0-π)x²(1－cos2x)dx＝1/2∫(0-π)x²dx－1/2∫(0-π)x²cos2xdx＝x³/6|(0-π)－1/16∫(0-π)(2x)²cos2xd(2x)＝π³/6－1/16∫(0-π)(2x)²cos2xd(2x)
记2x＝u，则
原式＝π³/6－1/16∫(0-2π)u²cosudu＝π³/6－1/16∫(0-2π)u²dsinu＝π³/6－1/16[u²sinu|(0-2π)－∫(0-2π)sinudu²]＝π³/6－1/16[(2π)²sin2π－0－2∫(0-2π)usinudu]＝π³/6－1/8∫(0-2π)udcosu＝π³/6－1/8[ucosu|(0-2π)－∫(0-2π)cosudu]＝π³/6－1/8[2πcos2π－0－sinu|(0-2π)]＝π³/6－π/4－sin2π＋sin0＝π³/6－π/4

∫(π/3-π/4)x/sinx²dx＝1/2∫(π/3-π/4)1/(sinx²)dx²
记x²＝u，则
原式＝1/2∫(π²/9-π²/16)cscudu＝1/2∫(π²/9-π²/16)cscu(cscu－cotu)/(cscu－cotu)du＝1/2∫(π²/9-π²/16)(－cscucotu＋csc²u)/(cscu－cotu)du＝1/2∫(π²/9-π²/16)1/(cscu－cotu)d(cscu－cotu)＝1/2ln|cscu－cotu| |(π²/9-π²/16)＝1/2(ln|cscπ²/16－cotπ²/16|－ln|cscπ²/9－cotπ²/9|)＝1/2ln|(cscπ²/16－cotπ²16)/(cscπ²/9－cotπ²/9)|

∫(0-π²/4)sin(x^1/2)dx
记x^1/2＝t，则x＝t²，有
原式＝∫(0-π/2)sintdt²＝2∫(0-π/2)tsintdt＝－2∫(0-π/2)tdcost＝－2[tcost|(0-π/2)－∫(0-π/2)costdt]＝－2[π/2·cosπ/2－0－sint(0-π/2)]＝2(sinπ/2－0)＝2