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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

这里给出的是拆分的方法... 用到cscx和cotx的原函数公式 请见下图

令t=sinx 原式=∫ t²dt =1/3t³+C 再把t=sinx带入 =1/3sin³x+C

x²-4≥0,x²≥4,设x=2secu,x²=4sec²u dx=2secutanudu sec²u=1+tan²u,sec²u-1=tan²u 换元得: ∫2tanu.2secutanudu =4∫secutan²udu =4∫tanudsecu =4secutanu-4∫secudtanu =4secutanu-4∫sec³udu...

∫[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 -...

你好 ∫x^2sin2xdx =-1/2∫x^2d(cos2x) =-1/2[cos2x*x^2-∫2x*cos2xdx] =-1/2[cos2x*x^2-∫xd(sin2x)] =-1/2[cos2x*x^2-(sin2x*x-∫sin2xdx)] =-1/2cos2x*x^2+1/2sin2x*x-1/2∫sin2xdx =-1/2cos2x*x^2+1/2sin2x*x+1/4cos2x+C 【数学辅导团】为您...

∫cos(√x)dx 令√x=u,则dx/2√x=du,dx=2(√x)du=2udu, 原式=2∫ucosudu =2∫ud(sinu) =2[usinu-∫sinudu] =2(usinu+cosu)+C =2[(√x)sin(√x)+cos(√x)]+C ~~~~~~~~~~~~~~~~~~~~~~~~~ ∫√x(x+1)^2dx 令√x=t, 则dx=2tdt,带入 =∫t(t^2+1)^2*2tdt =∫2t^6+4t^4...

令(x+1)^1/3=t,x=t^3-1,dx=3t^2dt ∫dx/[1+(x+1)^1/3] =∫3t^2/(1+t)dt =3∫t^2/(1+t)dt =3∫(t^2-1+1)/(1+t)dt =3∫[t-1+1/(1+t)]dt =3/2t^2-3t+3ln(1+t)+C 令 x^(1/4)=u,则 x=u^4,dx=4u^3du ∫dx/[x^(1/2)+x^(1/4)] = ∫4u^3du/[u^2+u] = 4∫[u-1+1/(u...

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