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Question Number 21679 by Arnab Maiti last updated on 30/Sep/17
∫((secθ dθ)/(1−secθ))
∫secθdθ1−secθ
Answered by alex041103 last updated on 30/Sep/17
First we make the following transformations: ∫((secθ dθ)/(1−secθ))=∫((1/(cosθ))/(1−(1/(cosθ))))dθ= =∫((1/(cosθ))/(1−(1/(cosθ)))) ((cosθ)/(cosθ))dθ= =∫(1/(cosθ−1))dθ Also by the double angle indetity we have cosθ=1−2sin^2 ((θ/2)) ⇒∫(1/(cosθ−1))dθ=−∫(1/(sin^2 ((θ/2)))) (dθ/2) Let x=(θ/2)⇒dx=(dθ/2) ∫(1/(cosθ−1))dθ=∫−csc^2 x dx=I We know that (d/dx)(ctg x)=−csc^2 x ⇒∫((secθ)/(1−secθ))dθ=ctg((θ/2))+C
Firstwemakethefollowingtransformations:∫secθdθ1−secθ=∫1cosθ1−1cosθdθ==∫1cosθ1−1cosθcosθcosθdθ==∫1cosθ−1dθAlsobythedoubleangleindetitywehavecosθ=1−2sin2(θ2)⇒∫1cosθ−1dθ=−∫1sin2(θ2)dθ2Letx=θ2⇒dx=dθ2∫1cosθ−1dθ=∫−csc2xdx=IWeknowthatddx(ctgx)=−csc2x⇒∫secθ1−secθdθ=ctg(θ2)+C

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