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求不定积分1/sin^4xCos^2xDx

∫1/(sin^4xcos^4x)dx =∫16/sin^4(2x)dx =∫16csc^4(2x)dx =∫16[cot^(2x)+1]csc^2(2x)dx =-∫8[cot^2(2x)+1]dcot(2x) =-8/3cot^3(2x)-8cot(2x)+C

错在划线那一步,你把cos(x/2)的指数弄错了

∫1/(sin)^4dx=-cosx/[3*(sinx)^3]+(2/3)∫1/(sinx)^2dx ∫1/(sinx)^2dx=-cotx 所以1/(sin)^4dx==-cosx/[3*(sinx)^3]-cotx+c,谢谢

积化和差的四个公式: sina*cosb=(sin(a+b)+sin(a-b))/2 cosa*sinb=(sin(a+b)-sin(a-b))/2 cosa*cosb=(cos(a+b)+cos(a-b))/2 sina*sinb=-(cos(a+b)-cos(a-b))/2

=-∫csc²xdcotx =-∫cot²x+1dcotx =-cot³x/3-cotx+C

原式=∫[1-1/(2sin²x+cos²x)]dx =∫dx - ∫[sec²x/(2tan²x+1)]dx =∫dx - ∫1/[(√2tanx)²+1)]d(tanx) =∫dx - √2/2* ∫1/[(√2tanx)²+1)]d(√2tanx) =x - ∫dx - √2/2*arctan(√2tanx)+c

∫ 1/sin⁴x dx = ∫ csc⁴x dx = ∫ csc²x d(-cotx) = -cotxcsc²x + ∫ cotx d(csc²x) = -cotxcsc²x - 2∫ cot²x•csc²x dx = -cotxcsc²x + 2∫ cot²x d(cotx) = -cotxcsc²x + (2/3)cot...

第二个括号里是啥

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