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∫sin²x/Cos³xDx

∫sin²xcos⁵xdx =∫sin²x(1-sin²x)²·cosxdx =∫(sin⁶x-2sin⁴x+sin²x)d(sinx) =(1/7)sin⁷x -(2/5)sin⁵x+(1/3)sin³x +C =(15sin⁴x-42sin²x+35)sin³x/105 +C

原题是:∫cos2x/(cos²xsin²x)dx=? 解:原式=4∫cos2x/(sin2x)^2dx =2∫1/(sin2x)^2d(sin2x) =-2/(sin2x)+C 希望对你有点帮助!

求不定积分∫sin²xdx解:原式=∫[(1-cos2x)/2]dx=(1/2)x-(1/2)∫cos2xdx=(1/2)x-(1/4)∫cos2xd(2x)=(1/2)x-(1/4)sin2x+C关于∫sinⁿxdx有递推公式:∫sinⁿxdx=-(sinⁿֿ¹xcosx)/n+[(n-1)/n]∫sinⁿֿ²xd...

解: ∫tan³xsecxdx =∫(sin³x/cos⁴x)dx =∫[sinx(1-cos²x)/cos⁴x]dx =∫(sinx/cos⁴x)dx -∫(sinx/cos²x)dx =-∫(1/cos⁴x)d(cosx)+∫(1/cos²x)d(cosx) =-(-⅓)(1/cos³x) -(1/cosx) +C =...

∫xcos²xdx=∫x(1+cos2x)/2dx=1/2(∫xdx+∫xcos2xdx) =1/2(1/2x²+∫xcos2xdx) =1/2(1/2x²+1/2∫xdsin2x) =1/2(1/2x²+1/2(xsin2x-∫sin2xdx)) =1/2(1/2x²+1/2xsin2x+1/4cos2x)+C

显然1-sin²x=cos²x 那么 ∫(sinx)^4/ cos²xdx =∫ (1-cos²x)² / cos²xdx =∫ [1-2cos²x+(cosx)^4] / cos²x dx =∫ (1/cos²x -2 +cos²x) dx = tanx -2x +∫ cos²x dx = tanx -2x +∫ (0.5+0.5c...

∫sin³xcos²xcos³xdx =∫sinx·(1-cos²x)cos²xcos³xdx =∫(-cos⁷x+cos⁵x)sinxdx =∫(cos⁷x-cos⁵x)d(cosx) =⅛cos⁸x -(1/6)cos⁶x +C

公式:∫[0→π] xf(sinx) dx = (π/2)∫[0→π] f(sinx) dx ∫[0→π] x(sinx)⁶(cosx)⁴ dx 由公式: =(π/2)∫[0→π] (sinx)⁶(cosx)⁴ dx =(π/2)∫[0→π/2] (sinx)⁶(cosx)⁴ dx + (π/2)∫[π/2→π] (sinx)⁶(cosx)⁴...

y=∫[0:x²]cos³tdt =∫[0:x²]cos²t·costdt =∫[0:x²](1-sin²t)d(sint) =sint -⅓sin³t|[0:x²] =sin(x²)-⅓sin³(x²)-(sin0-⅓sin0) =sin(x²)-⅓sin³(x&#...

sin3x=sin(x+2x)=sinxcos2x+cosxsin2x=sinx(1-2sin²x)+cosx(2sinxcosx) =sinx-2sin³x+2sinxcos²x =sinx-2sin³x+2sinx(1-sin²x) =3sinx-4sin³x 同理可得:cos3x=4cos³x-3cosx ∴原式=(4sinx-4sin³x)/(4c...

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