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1/1+tAn^2x的不定积分

1/1+tan^2x的不定积分 =∫1/sec^2xdx =∫cos^2xdx =∫(1+co2x)/2dx =1/2 x+1/4sin2x+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

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

原式=-∫1/(1+cos^2x)d(cosx) =-arctan(cosx)+C

其实第一题挺有难度的。。。 ∫ arctan√[(a - x)/(a + x)] dx = x * arctan√[(a - x)/(a + x)] - ∫ x * (- 1/2) * 1/√(a² - x²) dx = x * arctan√[(a - x)/(a + x)] + (1/2)∫ x/√(a² - x²) dx = x * arctan√[(a - x)/(a + x)...

∫dx/(x²+x+1) =4∫dx/(4x²+4x+1+3) =4∫dx/[(2x+1)²+3] = 4/3∫dx/{[(2x+1)/√3]²+1} = 2/√3∫d[(2x+1)/√3]/{[(2x+1)/√3]²+1} =2arctan[(2x+1)/√3]/√3+C

∫1/(1+cos²x)dx =∫sec²x/(sec²x+1)dx =∫1/(tan²x+2) dtanx =1/√2 arctan(tanx/√2)+c

1+(cotx)^2 =(cscx)^2

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