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1/(sEC^2x·tAn^2x)的不定积分

∫ (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

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

楼主,原式1+tan^2x可化为:

详细步骤写在纸上了

sec²x-1=1/cos²x -1=(1-cos²x)/cos²x=sin²x/cos²x=tan²x

2tan x/(1-tan^2 x)

tanx/2-1/2lin|cos2x|

=1/cos2x一sin2x/cos2x =(1一sin2x)/cos2x =(1一2sinxcosx)/(con^x一sin^x) =(sin×一conx)/(sinx十cosX)

∫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

注意d(tanx)=(secx)^2 dx 所以在这里得到 ∫ (tanx)^3 (secx)^2 dx =∫ (tanx)^3 d(tanx) =1/4 *(tanx)^4 +C,C为常数

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