sec-x-dx-

Question Number 29176 by sinx last updated on 04/Feb/18

$$\int{sec}\:{x}\:{dx}=? \\ $$

Commented by abdo imad last updated on 08/Feb/18

another method let put I=∫ (dx/(cosx)) the ch.tan((x/2))=t give I= ∫ (1/((1−t^2 )/(1+t^2 ))) ((2dt)/(1+t^2 ))= ∫((2dt)/(1−t^2 ))= ∫ ( (1/(1+t))+(1/(1−t)))dt =ln∣1+t∣−ln∣1−t∣ +k=ln∣((1+t)/(1−t))∣+k so I=ln∣((1+tan((x/2)))/(1−tan((x/2))))∣ +k .=ln∣tan((x/2)+(π/4))∣ +k .

$${another}\:{method}\:\:{let}\:{put}\:{I}=\int\:\frac{{dx}}{{cosx}}\:{the}\:{ch}.{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}=\:\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\:\int\frac{\mathrm{2}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }=\:\int\:\left(\:\frac{\mathrm{1}}{\mathrm{1}+{t}}+\frac{\mathrm{1}}{\mathrm{1}−{t}}\right){dt} \\ $$$$={ln}\mid\mathrm{1}+{t}\mid−{ln}\mid\mathrm{1}−{t}\mid\:+{k}={ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid+{k}\:{so} \\ $$$${I}={ln}\mid\frac{\mathrm{1}+{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\mid\:+{k}\:.={ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\mid\:+{k}\:. \\ $$

Answered by mrW2 last updated on 04/Feb/18

∫sec x dx=∫(1/(cos x)) dx=∫((cos x)/(cos^2 x))dx =∫(1/(1−sin^2 x)) d(sin x) =∫(1/(1−t^2 ))dt=(1/2)∫((1/(1+t))+(1/(1−t)))dt =(1/2)[ln ∣1+t∣−ln ∣1−t∣]+C =(1/2)ln ∣((1+t)/(1−t))∣+C =(1/2)ln ∣((1+sin x)/(1−sin x))∣+C

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

Answered by $@ty@m last updated on 05/Feb/18

∫sec x×(((sec x+tan x))/((sec x+tan x)))dx =∫((sec^2 x+sec xtan x)/(sec x+tan x))dx =ln (sec x+tan x)+C

$$\int{sec}\:{x}×\frac{\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)}{\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)}{dx} \\ $$$$=\int\frac{\mathrm{sec}\:^{\mathrm{2}} {x}+\mathrm{sec}\:{x}\mathrm{tan}\:{x}}{\mathrm{sec}\:{x}+\mathrm{tan}\:{x}}{dx} \\ $$$$=\mathrm{ln}\:\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)+{C} \\ $$

Answered by A1B1C1D1 last updated on 05/Feb/18

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