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1/(sin^4xCos^4x)的不定积分

分子分母同除以 cos^4x

=∫sin^4x(1-sin^2x)dsinx上面就是将sinx作为自变量,你可设sinx=u则:=fu^4(1-u^2)du=f[u^4-u^6]du 公式:(u^n)'=(n-1)^(n-1) ; fu^ndu=1/(n+1) *u^(n+1)+c=fu^4du-fu^6du=1/5u^5-1/7u^7+c再将u=sinx代入=1/5sin^5x-1/7sin^7x+c

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

∫ (sin²x - cos²x)/(sin⁴x + cos⁴x) dx = ∫ [- (cos²x - sin²x)]/[(sin⁴x + 2sin²xcos²x + cos⁴x) - 2sin²xcos²x] dx = ∫ (- cos2x)/[(sin²x + cos²x)² - 2sin...

前面是sinx的4次方还是sin4x啊

∫ 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...

错在划线那一步,你把cos(x/2)的指数弄错了

∫(sinx)^2*(cosx)^4dx =(1/4)∫(sin2x)^2(1-(sinx)^2)dx =(1/4)∫(sin2x)^2(1/2+cos2x/2)dx =(1/16)∫(1-cos4x)dx+(1/16)∫(sin2x)^2dsin2x =(1/16)x-(1/64)sin4x+(1/48)(sin2x)^3+C

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