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∫1/(2+sinx)Dx

2+sinx=2sin(x/2)^2+2cos(x/2)^2+2sin(x/2)cos(x/2) dx/(2+sinx)=sec(x/2)^2dx/[2+2tan(x/2)^2+2tan(x/2)] =d(tan(x/2))/[1+tan(x/2)+tan(x/2)^2] 令u=tan(x/2) 原积分=∫du/(1+u+u^2) =∫d(u+1/2)/[3/4+(u+1/2)^2](用∫dx/(a^2+x^2)公式,取a=√3/...

令t=tan(x/2),则x=2arctant,所以dx=2/(1+t^2)dt 由万能公式:sinx=2tan(x/2)/(1+(tan(x/2))^2)=2t/(1+t^2), 则原式=(1/2)∫d(t+1/2)/[(t+1/2)^2+(根号3/2)^2] =(1/根号3)arctan[2(t+1/2)/根号3]+C =(1/根号3)arctan[2(arctan(x/2)+1/2)/根号3]+C

转换方法:

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

解:分子分母同除以(cosx)^2得: 然后套公式:

解:分享一种解法。 将积分区间[0,2π]拆成[0,π/2)∪[π/2,π)∪[π,3π/2)∪[3π/2,2π),则∫(0,2π)dx/(2+sinx)=∫(0,π/2)dx/(2+sinx)+∫(π/2,π)dx/(2+sinx)+∫(π,3π/2)dx/(2+sinx)+∫(3π/2,2π)dx/(2+sinx),对后三个积分,分别设x=t+π/2、t+π、t+3π/2,则 ∴∫...

令u=tan(x/2),则sinx=2u/(1+u^2),dx=2du/(1+u^2) 原式=∫1/[1+4u/(1+u^2)]*2du/(1+u^2) =∫2du/(1+u^2+4u) =∫2du/[(u+2)^2-3] =∫2du/(u+2+√3)(u+2-√3) =(√3/3)*[∫du/(u+2-√3)-∫du/(u+2+√3)] =(√3/3)*[ln|u+2-√3|-ln|u+2+√3|]+C =(√3/3)*[ln|tan(...

这个是三角函数的不定积分,分母应先进性化简,计算步骤为: ∫1/(sinx+cosx)dx =∫dx/√2sin(x+π/4) =-(√2/2)∫dcos(x+π/4)/sin^2(x+π/4) =-(√2/4){∫dcos(x+π/4)/[1-cos(x+π/4)]+∫dcos(x+π/4)/[1+cos(x+π/4)]} =-(√2/4)ln{[1+cos(x+π/4)]/[1-cos...

三角代换 过程如下图:

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