# Integration by substitutionWatch

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#1
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Thanks
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13 years ago
#2
u=tanx, du/dx=(secx)^2=1+(tanx)^2=1+u^2, dx=1/(1+u^2) du
x=pi/6, u=1/rt3, x=0, u=0.
∫(1+tanx)/(1-tanx) dx with limits 0, pi/6.
=∫(1+u)/(1-u).1/(1+u^2) du with limits 0, 1/rt3.
=∫(1+u)/(1-u)(1+u^2) du
(1+u)/(1-u)(1+u^2)=A/(1-u)+(Bu+C)/(1+u^2)
(1+u)=A(1+u^2)+(Bu+C)(1-u)
u=1, A=1.
u=0, 1=1+C, C=0.
u=2, 3=5-2B
B=1.
∫1/(1-u) + u/(1+u^2) du
= ∫1/(1-u) du + 0.5∫2u/(1+u^2) du, both with limits 0, 1/rt3.
=[ -ln[1-u] + 0.5ln[1+u^2] ]. Now you just need to evaluate the limits.
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#3
Thanks AGAIN!
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