sec-d-

Question Number 224833 by fantastic last updated on 06/Oct/25

$$\int\mathrm{sec}\:\theta\:{d}\theta \\ $$

Answered by taha3738 last updated on 06/Oct/25

∫ sec θ dθ = ∫ ((sec θ (sec θ + tan θ))/(sec θ + tan θ)) dθ = ∫ ((sec^2 θ+ sec θ tan θ)/(sec θ + tan θ )) dθ = ln ∣ sec x + tan x ∣ + C Because (d/dθ) ( sec θ + tan θ ) = (d/dθ) sec θ + (d/dθ) tan θ = sec θ tan θ + sec^2 θ

$$\int\:\mathrm{sec}\:\theta\:{d}\theta\:=\:\int\:\frac{\mathrm{sec}\:\theta\:\left(\mathrm{sec}\:\theta\:+\:\mathrm{tan}\:\theta\right)}{\mathrm{sec}\:\theta\:+\:\mathrm{tan}\:\theta}\:{d}\theta \\ $$$$=\:\int\:\:\frac{\mathrm{sec}^{\mathrm{2}} \theta+\:\mathrm{sec}\:\theta\:\mathrm{tan}\:\theta}{\mathrm{sec}\:\theta\:+\:\mathrm{tan}\:\theta\:}\:{d}\theta\:=\:\mathrm{ln}\:\mid\:\mathrm{sec}\:{x}\:+\:\mathrm{tan}\:{x}\:\mid\:+\:{C} \\ $$$${Because}\:\frac{{d}}{{d}\theta}\:\left(\:\mathrm{sec}\:\theta\:+\:\mathrm{tan}\:\theta\:\right)\:=\:\frac{{d}}{{d}\theta}\:\mathrm{sec}\:\theta\:+\:\frac{{d}}{{d}\theta}\:\mathrm{tan}\:\theta \\ $$$$\:=\:\mathrm{sec}\:\theta\:\mathrm{tan}\:\theta\:+\:\mathrm{sec}^{\mathrm{2}} \:\theta\: \\ $$

Commented by fantastic last updated on 06/Oct/25

$${thank}\:{you}\:{very}\:{much}\:! \\ $$

Answered by mehdee7396 last updated on 06/Oct/25

∫(dθ/(cosθ))=∫((1+tan^2 (θ/2))/(1−tan^2 (θ/2)))dθ→^(tan(θ/2)=u) ∫((2du)/(1−u^2 ))=ln((1+u)/(1−u))+c=ln((1+tan(θ/2))/(1−tan(θ/2)))+c =ln((cos(θ/2)+sin(θ/2))/(cos(θ/2)−sin(θ/2)))+c=ln((1+sinθ)/(cosθ))+c

$$\int\frac{{d}\theta}{{cos}\theta}=\int\frac{\mathrm{1}+{tan}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{1}−{tan}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{d}\theta\overset{{tan}\frac{\theta}{\mathrm{2}}={u}} {\rightarrow} \\ $$$$\int\frac{\mathrm{2}{du}}{\mathrm{1}−{u}^{\mathrm{2}} }={ln}\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}+{c}={ln}\frac{\mathrm{1}+{tan}\frac{\theta}{\mathrm{2}}}{\mathrm{1}−{tan}\frac{\theta}{\mathrm{2}}}+{c} \\ $$$$={ln}\frac{{cos}\frac{\theta}{\mathrm{2}}+{sin}\frac{\theta}{\mathrm{2}}}{{cos}\frac{\theta}{\mathrm{2}}−{sin}\frac{\theta}{\mathrm{2}}}+{c}={ln}\frac{\mathrm{1}+{sin}\theta}{{cos}\theta}+{c} \\ $$$$ \\ $$

Commented by fantastic last updated on 06/Oct/25

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Answered by Ghisom_ last updated on 07/Oct/25

∫(dθ/(cos θ))= [t=((1+sin θ)/(cos θ)) → dx=((cos^2 x)/(1+sin x))dt=((cos x)/t)dt] =∫(dt/t)=ln ∣t∣ =ln ∣((1+sin θ)/(cos θ))∣ +C

$$\int\frac{{d}\theta}{\mathrm{cos}\:\theta}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{1}+\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\:\rightarrow\:{dx}=\frac{\mathrm{cos}^{\mathrm{2}} \:{x}}{\mathrm{1}+\mathrm{sin}\:{x}}{dt}=\frac{\mathrm{cos}\:{x}}{{t}}{dt}\right] \\ $$$$=\int\frac{{dt}}{{t}}=\mathrm{ln}\:\mid{t}\mid\:=\mathrm{ln}\:\mid\frac{\mathrm{1}+\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\mid\:+{C} \\ $$

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