求∫(0-π/2)e^2xcosxdx,∫(0-π)(xsinx)^2dx,∫(π/3-π/4)x/sinx^2dx,∫
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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
![求∫(0-π/2)e^2xcosxdx,∫(0-π)(xsinx)^2dx,∫(π/3-π/4)x/sinx^2dx,∫](/uploads/image/z/7940847-39-7.jpg?t=%E6%B1%82%E2%88%AB%280-%CF%80%2F2%29e%5E2xcosxdx%2C%E2%88%AB%280-%CF%80%29%28xsinx%29%5E2dx%2C%E2%88%AB%28%CF%80%2F3-%CF%80%2F4%29x%2Fsinx%5E2dx%2C%E2%88%AB)
∫(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
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
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