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求sEC^2xtAn^3x的不定积分

∫sec^3x*tan^3xdx =∫sec^2xtan^2x dsecx =∫sec^2x(sec^2x-1) dsecx =∫(sec^4x-sec^2x) dsecx =1/5sec^5x-1/3sec^3x+c

∫ (tan2x + sec2x)² dx = ∫ (tan²2x + 2sec2xtan2x + sec²2x) dx = (1/2)∫ (sec²2x - 1 + 2sec2xta2x + sec²2x) d(2x) = (1/2)(2tan2x - 2x + 2sec2x) + C = tan2x + sec2x - x + C

∫tan^2xsec^4xdx =∫tan^2xsec^2xd(tanx) =∫tan^2x(tan^2x+1)d(tanx) =∫(tan^4x+tan^2x)d(tanx) =(1/5)tan^5x+(1/3)tan^3x+C

这是公式 ∫ (sec^2x - csc^2x) dx = ∫ sec^2x dx - ∫ csc^2x dx = tanx + cotx + c 欢迎采纳,不要点错答案哦╮(╯◇╰)╭

∫sec^2xtanxdx=∫tanxdtanx=(1/3)tan^3x+C

∫sec²x dx =∫d(tanx) =tanx+C 这个是基本积分公式之一,必须记好 因为d/dx (tanx)=sec²x

∫tan³xsec³xdx 解法一: ∫tan³xsec³xdx =∫ tan²x sec²x dsecx =∫(sec²x - 1) * sec²x dsecx =∫( sec^4x - sec²x ) dsecx = sec^5x / 5 - sec^3x / 3 + C 解法二 ∫tan³x sec³xdx ...

原式=∫ (1/(2^cos^x-1))*sinx/cosx dx = - ∫ (1/(2^cos^x-1)) / cosx d(cosx) 设u=cosx,化简得 原式=- ∫ 1/((2u^2-1)*u)du =-∫ 根号2/2(1/(根号2u-1)+1/(1+根号2u))+1/udu =-1/2(ln(根号2u-1)+ln(根号2u+1))+lnu+C =ln|cosx|-1/2*ln|cos2x|+C

∫tan²x/(1-sin²x) dx =∫tan²x/cos²x dx =∫tan²x*sec²x dx =∫tan²x d(tanx) =(1/3)tan³x + C

∫[tanx/(secx)^2dx =∫(sinx/cosx)(cosx)^2dx =∫sinxcosx =∫sinxd(sinx) =(1/2)(sinx)^2+C。

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