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∫xtAnx sECx 4Dx

用凑微分 解:∫xtanx(secx)^4dx=∫xtanx([(tanx)^2+1]dtanx=∫x(tanx)^3dtanx+∫xtanxdtanx=(1/4)∫xd(tanx)^4+(1/2)∫xd(tanx)^2=(1/4)[x(tanx)^4-∫(tanx)^4dx]+(1/2)[x(tanx)^2-∫(tanx)^2dx](*) 因为 ∫(tanx)^4dx=∫(tanx)^2((secx)^2-1)dx=∫(tanx)^2dtanx-∫

原式=∫(secx)/(1+tanx)dx =∫1/(1+tanx)d(tanx) =∫1/(1+tanx)d(1+tanx) =[-1/(1+tanx)]+c

∫(secx)^4dx=tanx+1/3*(tanx)^3 +C.C为常数.解答过程如下:∫(secx)^4dx =∫(secx)^4dx=∫(secx)^2*(secx)^2dx=∫(1+(tanx)^2)*(1+(tanx)^2)dx 令y=tanx,则dy=(1+(tanx)^2)dx=(1+y^2)dx 上式=∫(1+y^2)dy=y+1/3*y^3=tanx+1/3*(tanx)^3 +C 扩展资料:

∫xtanxdtanx =x(tanx)^2 - ∫tanx ( tanx + x(secx)^2 dx=x(tanx)^2 - ∫ [(secx)^2 -1 + (1/2)x.sin2x ] dx =x(tanx)^2 - tanx + x -(1/2)∫x.sin2x dx=x(tanx)^2 - tanx + x +(1/4)∫x dcos2x =x(tanx)^2 - tanx + x +(1/4)x.cos2x -(1/4) ∫cos2x dx=x(tanx)^2 - tanx + x +(1/4)x.cos2x -(1/8) sin2x + C

分部积分法:xtanx(secx^4)dx=xsec^3xdsecx=xdsec^4x/4=x*sec^4x/4-sec^4/xdx=x*sec^4x/4-1/4(1+tan^2x)dtanx=x*sec^4x/4-1/4tanx-1/4*tan^3x/3=1/4(x*sec^4x-tanx-tan^3x/3)

(tanx)^4 ?(tanx)^2=(secx)^2 - 1∫(tanx)^4dx=∫(tanx)^2*[(secx)^2-1]dx=∫(tanx)^2*d(tanx)-∫(tanx)^2dx=1/3*(tanx)^3-∫[(secx)^2-1]dx=1/3*(tanx)^3-tanx+x+c

∫(tanx)^4 dx=∫(tanx)^2 *[(secx)^2 -1] dx=∫(tanx)^2 * (secx)^2 dx - ∫(tanx)^2 dx=∫(tanx)^2 d(tanx) - ∫[(secx)^2 -1 ]dx=(tanx)^3/3 - ∫(secx)^2 dx +∫ 1 dx=(tanx)^3/3 -tanx + x + C

integral((tan(x))^3*(sec(x))^4,x)-1/12*(3*sin(x)^2 - 1)/(sin(x)^6 - 3*sin(x)^4 + 3*sin(x)^2 - 1)

原式=∫(secx)^4dx=∫(secx)^2*(secx)^2dx=∫(1+(tanx)^2)*(1+(tanx)^2)dx 令y=tanx,则dy=(1+(tanx)^2)dx=(1+y^2)dx 上式=∫(1+y^2)dy=y+1/3*y^3=tanx+1/3*(tanx)^3 +C

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