f(1/sinxcosx)dx

∫dx/(1+sinxcosx)^2∫dx/(1+sinxcosx)^2∫dx/(1+sinxcosx)^2令v=tanx,dx=dv/(1+v^2),sinx=v/√(1+v^2),cosv=1/√

∫sinxcosx/(1+sin^4x)dx∫sinxcosx/(1+sin^4x)dx∫sinxcosx/(1+sin^4x)dx∫sinxcosx/(1+sin^4x)dx=∫sinx/(1+si

求不定积分sinxcosx/1+sinx^4dx求不定积分sinxcosx/1+sinx^4dx求不定积分sinxcosx/1+sinx^4dx∫sinxcosx/(1+sin^4x)dx=∫sinx

∫sinxcosx/[1+(sinx)^4)]dx∫sinxcosx/[1+(sinx)^4)]dx∫sinxcosx/[1+(sinx)^4)]dx用第一换元法（sinxcosx=1／2d（sin^

∫(1/sinxcosx)dx求定积分∫(1/sinxcosx)dx求定积分∫(1/sinxcosx)dx求定积分∫(1/sinxcosx)dx=∫1/sinx*d(sinx)=1/2(sinx)^2

∫cos2x/(1+sinxcosx)dx求详解.∫cos2x/(1+sinxcosx)dx求详解.∫cos2x/(1+sinxcosx)dx求详解.Letu=1+sin(x)cos(x)=1+(1/

∫（sinxcosx）/（1+sinx^4）dx求导?∫（sinxcosx）/（1+sinx^4）dx求导?∫（sinxcosx）/（1+sinx^4）dx求导?∫[(sinxcosx)/（1+sin

积分sinxcosx/(1+sin^2x)^5dx积分sinxcosx/(1+sin^2x)^5dx 积分sinxcosx/(1+sin^2x)^5dx全部展开收起原式=1/2∫1/(1+s

求解∫(x+2sinxcosx)／(1+cos2x)dx求解∫(x+2sinxcosx)／(1+cos2x)dx求解∫(x+2sinxcosx)／(1+cos2x)dx∫(x+2sinxcosx)/(

∫cosx^2/1+sinxcosxdx如何求解∫cosx^2/1+sinxcosxdx如何求解∫cosx^2/1+sinxcosxdx如何求解∫cos²x/(1+sinxcosx)dx=∫

∫（COS2X）／（1十SinXCOSX）dX＝∫（COS2X）／（1十SinXCOSX）dX＝∫（COS2X）／（1十SinXCOSX）dX＝∫（COS2X）／（1十SinXCOSX）dX=∫(1/

∫sinxcosx/(sinx+cosx)dx∫sinxcosx/(sinx+cosx)dx∫sinxcosx/(sinx+cosx)dx∫sin2xdx/(sinx+cosx)=∫cos(π/2-2

求不定积分∫dx/(sinxcosx)求不定积分∫dx/(sinxcosx)求不定积分∫dx/(sinxcosx)

∫cos2x/(sinxcosx)^2dx∫cos2x/(sinxcosx)^2dx∫cos2x/(sinxcosx)^2dx∫cos2x/(sinxcosx)^2dx=∫√2cos2x/(2sinx

化简f(x)=cos2x+sinxcosx+1化简f(x)=cos2x+sinxcosx+1化简f(x)=cos2x+sinxcosx+1sinxcosx=1/2sin2x,所以f(x)=cos2x+

1.∫（sinxcosx）/（1+sin^2x）dx2.设f''（sinx）=cos^2x则f(x)=3.已知f''(cosx)=sinx则f(cosx)=4.∫dx/(e^x-1)^1/2=.....第

已知f(tanx)=1/sinxcosx,求f(x)=已知f(tanx)=1/sinxcosx,求f(x)=已知f(tanx)=1/sinxcosx,求f(x)=原式＝sin2x+cos2x/sinx

f(x)=cosx-根号3sinxcosx+1化简（sinxcosx不在根号内）f(x)=cosx-根号3sinxcosx+1化简（sinxcosx不在根号内）f(x)=cosx-根号3sinxcos

计算不定积分1.∫(sinXcosx)/(1+sin^4X)dx2.∫dx/(X^2(4-X^2)^1/2)计算不定积分1.∫(sinXcosx)/(1+sin^4X)dx2.∫dx/(X^2(4-X

计算不定积分1.∫(sinXcosx)/(1+sin^4X)dx2.∫dx/(X^2(4-X^2)^0.5计算不定积分1.∫(sinXcosx)/(1+sin^4X)dx2.∫dx/(X^2(4-X^