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∫(1+(Cosx)^(1/2))DCosx

答案在图片上,希望得到采纳,谢谢。愿您学业进步☆⌒_⌒☆

∫1/(sinx)dx =∫cscxdx =∫sinx/(1-cos²x) dx =-∫dcosx/(1-cos²x) =-1/2[∫dcosx/(1-cosx)+∫dcosx/(1+cosx)] = -1/2[∫-d(1-cosx)/(1-cosx)+∫d(1+cosx)/(1+cosx)] =-1/2ln(1+cosx)/ (1-cosx)+C =ln[(1-cosx)/sinx]+C =ln(cscx-cotx)+C

等下给你答案啊

先求不定积分 ∫1/sinx dx =∫sinx/sin²xdx =-∫1/sin²xdcosx =-∫1/(1-cos²x)dcosx =∫1/(cosx+1)(cosx-1)dcosx =∫[1/(cosx-1)-1/(cosx+1)]/2dcosx =[∫1/(cosx-1)dcosx-∫1/(cosx+1)dcosx]/2 =[∫1/(cosx-1)d(cosx-1)-∫1/(cosx+1)d(cos...

令cosx=t ∫ [1/(1-t)]dt=∫ [1/(1-t)]d(1-t)=-ln(1-t)+C 所以 ∫ [1/(1-cosx)]dcosx=-ln(1-cosx)+C 【数学之美】很高兴为你解答,不懂请追问!满意请采纳,谢谢!O(∩_∩)O~

见图

cosx=1-2sin²(x/2)∫(2-2cosx)^0.5dx=∫[2-2+4sin²(x/2)]^0.5 dx=∫ 2|sin(x/2)| dx①当2kπ≤x/2≤π+2kπ时,原式=∫2sin(x/2)dx=4∫sin(x/2)d(x/2)=4cos(x/2)+C②当π+2kπ≤x/2≤2π+2kπ时,原式=-∫2s...

f'(cosx)=sinx f'(cosx)dcosx=sinxdcosx 积分,得 f(cosx)=∫sinxdcosx =-∫sin²xdx =-1/2∫(1-cos2x)dx =-1/2x+1/4sin2x+c

“cosx”为微元啊! ∫sin³xdx =-∫(1-cos²x)d(cosx) =-cosx+(1/3)cos³x+C。

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