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高数求不定积分∫1/(4Cos^2x+9sin^2x)Dx

亲 麻烦点一下采纳谢谢

还需要帮忙的话可以先采纳再详解

原式=∫sin²x (sinxcosx)² dx =∫sin²x * (sin2x / 2)² dx =1/4 ∫(1 - cos2x)/2 * ( 1 - cos4x)/2 dx =1/16 ∫(1 - cos2x) * ( 1 - cos4x) dx =1/16 ∫(cos4xcos2x - cos2x - cos4x + 1) dx =1/16 ∫((cos6x + cos2x) / 2 - cos2...

arctan(tanx)=x

原式=∫cos^3x/[√2*sin(x+π/4)]dx 令t=x+π/4,则dx=dt 原式=∫cos^3(t-π/4)/(√2*sint)dt =∫[√2/2*(cost+sint)]^3/(√2*sint)dt =(1/4)*∫(cost+sint)^3/sintdt =(1/4)*∫(cos^3t+3cos^2tsint+3costsin^2t+sin^3t)/sintdt =(1/4)*∫[cost(1-sin^2t)/sin...

∫[1/(1+x^4)]dx = 1/2∫[(x^2+1)-(x^2-1)]/(1+x^4)dx = 1/2 {∫(x^2+1)/(1+x^4) dx - ∫(x^2-1)/(1+x^4)dx } = 1/2 {∫(1+1/x^2)dx /(x^2+1/x^2) - ∫(1-1/x^2)dx/(x^2+1/x^2)} = 1/2 {∫d(x-1/x) /[(x-1/x)^2+2] - ∫d(x+1/x) /[(x+1/x)^2 -...

其实就是将1-x²看作整体

2e^(2x) +2e^x +1 =2(e^x + 1/2)^2 + 1/2 let e^x + 1/2 = (1/2)tanu e^x dx =(1/2)(secu)^2 du dx =(1/2)(secu)^2 du /[ (1/2)tanu -1/2 ] = (secu)^2 du / ( tanu -1 ) ∫dx/√(2e^(2x) +2e^x +1) =∫[(secu)^2 du / ( tanu -1 )]/[ (secu)/√2] =...

正弦的倒数为余割 sinx=1/cscx

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