急～ㄧ題不定積分問題

利用半角代換法
令z=tan(x/2)
則sinx=2z/(1+z^2)
cosx=(1-z^2)/(1+z^2)
dx=2dz/(1+z^2)
則所求=2∫[1+z^2/(z^2+2z-1)]*[1/(1+z^2)]dz
=2∫dz/(z^2+2z-1)
=2∫dz/[(z+1)^2-2]
令z+1=√2(secθ)
dz=√2secθtanθdθ
則原式=∫secθtanθ/(tan^2θ)dθ
=∫(1/sinθ)dθ
=ln lcscθ-cotθl +C
再將θ變換回去即可

算錯請指正

2010-01-13 11:41:33 補充：
∫tanx/(sec^2x+4)dx
=∫sinxcosx/(1+4cos^2x)dx
=∫sinxcosx/(1+2+2cos2x)dx
=(1/4)∫sin2x/(3+2cos2x)d2x
利用半角代換法
令z=tanx
sin2x=2z/(1+z^2)，cos2x=(1-z^2)/(1+z^2)
d2x=2dz/(1+z^2)
則所求=∫[z/(5+z^2)(1+z^2)]dz
=(1/4)[∫zdz/(1+z^2)-∫zdz/(5+z^2)]
=(1/8)[ln(1+z^2)-ln(5+z^2)]+C
=(1/8)[ln(sec^2x)-ln(5+tan^2x)]+C

算錯請指正

1.
∫1/(sinx-cosx)dx= 1/√2 *∫csc(x-π/4) dx
=1/√2 * ln| csc(x-π/4) - cot(x-π/4)|+C
= 1/√2 * ln| (√2-sinx-cosx)/(sinx-cosx)| +C

2.
∫ tanx/[(secx)^2+4] dx (同乘以(cosx)^2
=∫sinx*cosx/[1+4(cosx)^2] dx (sub. u=1+4(cosx)^2)
=(-1/8)∫ du/u
=(-1/8) ln[ 1+4(cosx)^2] + C

Stanx/(sec^2x+4)dx=S(sinxcosx)/(1+4cos^2x)dx
u=cosx du=-sinxcosx
=-S u/(1+4u^2)du=-1/8ln(1+4u^2)=-1/8ln(1+4cos^2x)

2010-01-13 14:59:48 補充：
du=-sinxdx