急!!!求下列不定积分
∫tanx(tanx-secx)dx
=∫(tan^2 x-tanxsecx)dx
=∫tan^2 xdx-∫tanxsecxdx
=∫(sec^2 x-1)dx-secx
=∫sec^2 xdx-x-secx
=tanx-x-secx+C
∫sin(ωt+θ)dt
=(1/ω)∫sin(ωt+θ)d(ωt+θ)
=(-1/ω)*cos(ωt+θ)+C
∫cos2x/(sin^2 x*cos^2 x)dx
=∫(cos^2 x-sin^2 x)/(sin^2 x*cos^2 x)dx
=∫[(1/sin^2 x)-(1/cos^2 x)]dx
=∫(csc^2 x-sec^2 x)dx
=...全部
∫tanx(tanx-secx)dx
=∫(tan^2 x-tanxsecx)dx
=∫tan^2 xdx-∫tanxsecxdx
=∫(sec^2 x-1)dx-secx
=∫sec^2 xdx-x-secx
=tanx-x-secx+C
∫sin(ωt+θ)dt
=(1/ω)∫sin(ωt+θ)d(ωt+θ)
=(-1/ω)*cos(ωt+θ)+C
∫cos2x/(sin^2 x*cos^2 x)dx
=∫(cos^2 x-sin^2 x)/(sin^2 x*cos^2 x)dx
=∫[(1/sin^2 x)-(1/cos^2 x)]dx
=∫(csc^2 x-sec^2 x)dx
=∫csc^2 xdx-∫sec^2 xdx
=-cotx-tanx+C
∫(cos√x)/√xdx
令√x=t,则x=t^2,那么dx=d(t^2)=2tdt
原式=∫[(cost)/t]*2tdt
=2∫costdt
=2sint+C
=2sin√x+C
∫(t-1)^2/√tdt
令√t=x,则t=x^2,那么dt=d(x^2)=2xdx
原式=∫[(x^2-1)^2/x]*2xdx
=2∫(x^2-1)^2dx
=2∫(x^4-2x^2+1)dx
=2*[(1/5)x^5-(2/3)x^3+x]+C
=(2/5)*t^(5/2)-(4/3)*t^(3/2)+2√t+C
∫(x+1)/√(1-x^2)dx
令x=sinθ,则dx=d(sinθ)=cosθdθ
原式=∫[(sinθ+1)/√(1-sin^2 θ)]*cosθdθ
=∫[(sinθ+1)/cosθ]*cosθdθ
=∫(sinθ+1)dθ
=∫sinθdθ+θ
=-cosθ+θ+C
=-√(1-x^2)+arcsinx+C
∫(x+1)^2/[x(x^2+1)]dx
=∫(x^2+2x+1)/[x(x^2+1)]dx
=∫[(x^2+1)+2x]/[x(x^2+1)]dx
=∫(1/x)dx+∫2/(x^2+1)dx
=ln|x|+2∫1/(x^2+1)dx
=ln|x|+2arctanx+C
∫1/[x(x^4+1)]dx
令x=1/t,则t=1/x,那么dx=d(1/t)=(-1/t^2)dt
原式=∫1/[(1/t)*(1/t^4+1)]*(-1/t^2)dt
=-∫1/[(1/t^4)+1]*(1/t)dt
=-∫[t^4/(t^4+1)]*(1/t)dt
=-∫[t^3/(t^4+1)]dt
=-(1/4)∫[1/(t^4+1)]d(t^4+1)
=-(1/4)ln|t^4+1|+C
=(-1/4)ln|(1/x^4)+1|+C。
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