∫cos2x/(cos^2xsin^2x)dx

∫cos2x/(cos^2xsin^2x)

1/(sin^2xcos^2x)=2/sin^2(2x)=2csc^2(2x) 而(cotx)'=-csc^2x 易得所求为-cot2x+c

求积分 cos2x/cos^xsin^2x dx

cos2x / (sinx * cosx) dx = ∫ cos2x / (1/2 * sin2x) dx = 4∫ cos2x / (sin2x) dx = 4∫ csc2x * cot2x dx = -2∫ csc2x * cot2x d(2x) = -2csc2x + C = -2/

请问cos2x/(cos^2xsin^2x)怎么算?

=∫(cos^2x-sin^2x)/(cos^2xsin^2x)=∫(1/sin^2x - 1/cos^2x)=∫csc^2x-∫sec^2x=cotx-tanx

∫cos2x/sin^2xcos^2x dx

∫cos2x/sin^2xcos^2x dx= ∫cos2x/(1/2sin2x)^2 dx=∫2/(sin2x)^2 dsin2x 再利用换元积分得=lnsin2x lnsin2X

∫(cos2x/sin^2x)dx

1.将分母变为sin2x即原式为∫[(4cos2x/sin^2(2x))]dx2.进行换元即2x变为t,原式变为∫[(2cos2x/sin^2t)]dt.3继续换元,可观察到(sin t)'=cost.所以原式等于2∫[(1/sin^2t]d(sint).4.得出答案为:(-2/sint)+c5.将t换回为2x有(-2/sin2x)+c.手打很累,望采纳.

求∫cos2x/cos^xsin^dx 的不定积分

|∫=∫(sin²x+cos²x)/(cos²xsin²x)dx=∫(1+tan²x)/(sin²x)dx=-∫dcot²x+∫dcot²x/cot²x=-cot²x+ln|cot²x|+C

∫(cos2x/sin^2x)dx

cos2x=cos*2x-sin*2x cot*2x=csc*2x-1 cotx的导数=-csc*2x∫(cos2x/sin^2x)dx=∫cos*2x-sin*2x/sin^2xdx 经过化简得 =∫cot*2xdx-∫1dx =∫csc*2xdx-∫1dx-∫1dx =-cotx-2x+c

跪求∫cos2x/sin^2xcos^2xdx 十万火急!!

sin^2xcos^2x=1/4(2sinxcosx)^2=1/4sin2x 原式=∫cos2x/sin^2xcos^2xdx=∫cos2x/(1/4sin2x)dx=∫2/sin2xd(sin2x)=2lnsin2x+C

求不定积分 ∫(cos2x/cos^2x)dx

∫ cos2x / (sin²x * cos²x) dx= ∫ cos2x / (1/2 * sin2x)² dx= 4∫ cos2x / (sin²2x) dx= 4∫ csc2x * cot2x dx= -2∫ csc2x * cot2x d(2x)= -2csc2x + c= -2/(sin2x) + c= -secx*cscx + c