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Question Number 38397 by Fawomath last updated on 25/Jun/18

evaluate ∫secxdx

$${evaluate}\:\:\int{secxdx} \\ $$

Commented by maxmathsup by imad last updated on 25/Jun/18

let I= ∫ (dx/(cosx)) changement 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)/(1−t))∣+c=ln∣ ((1+tan((x/2)))/(1−tan((x/2))))∣+c =ln∣tan((π/4)+(x/2))∣ +c

$${let}\:{I}=\:\int\:\:\:\frac{{dx}}{{cosx}}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}=\:\:\int\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\:=\:\int\:\left(\:\frac{\mathrm{1}}{\mathrm{1}+{t}}\:+\frac{\mathrm{1}}{\mathrm{1}−{t}}\right){dt} \\ $$$$={ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid+{c}={ln}\mid\:\frac{\mathrm{1}+{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\mid+{c}\:={ln}\mid{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)\mid\:+{c} \\ $$

Commented by Cheyboy last updated on 25/Jun/18

∫ secx dx ∫ (secx .((secx+tanx)/(secx+tanx)))dx ∫ (((sec^2 x+secxtanx)/(secx+tanx))) dx let u=secx+tanx du=(secxtanx+sec^2 x)dx ∫(1/u) du ln∣u∣ +C ln ∣secx+tanx∣+C

$$\int\:\boldsymbol{{secx}}\:\boldsymbol{{dx}} \\ $$$$\int\:\left(\boldsymbol{{secx}}\:.\frac{\boldsymbol{{secx}}+\boldsymbol{{tanx}}}{\boldsymbol{{secx}}+\boldsymbol{{tanx}}}\right)\boldsymbol{{dx}} \\ $$$$\int\:\left(\frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}+\boldsymbol{{secxtanx}}}{\boldsymbol{{secx}}+\boldsymbol{{tanx}}}\right)\:\boldsymbol{{dx}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{u}}=\boldsymbol{{secx}}+\boldsymbol{{tanx}} \\ $$$$\boldsymbol{{du}}=\left(\boldsymbol{{secxtanx}}+\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}\right)\boldsymbol{{dx}} \\ $$$$\int\frac{\mathrm{1}}{\boldsymbol{{u}}}\:\boldsymbol{{du}} \\ $$$$\mathrm{ln}\mid\boldsymbol{{u}}\mid\:+{C} \\ $$$$\mathrm{ln}\:\mid\boldsymbol{{secx}}+\boldsymbol{{tanx}}\mid+{C} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18

∫(dx/(cosx)) =∫((cosx)/(cos^2 x))dx =∫((cosx)/(1−sin^2 x))dx t=sinx dt=cosxdx ∫(dt/(1−t^2 ))=(1/2)∫((1+t+1−t)/((1+t)(1−t)))dt (1/2)∫(dt/(1−t))+(1/2)∫(dt/(1+t)) =−(1/2)ln∣1−t∣+(1/2)ln∣1+t∣ =(1/2)ln∣((1+t)/(1−t))∣+c =(1/2)ln∣((1+sinx)/(1−sinx))∣+c =(1/2)ln∣(((1+sinx)^2 )/(1−sin^2 x))∣+c =(1/2)ln∣(((1+sinx)^2 )/(cos^2 x))∣+c =ln∣(1/(cosx))+((sinx)/(cosx))∣+c =ln∣secx+tanx∣+c

$$\int\frac{{dx}}{{cosx}} \\ $$$$=\int\frac{{cosx}}{{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int\frac{{cosx}}{\mathrm{1}−{sin}^{\mathrm{2}} {x}}{dx} \\ $$$${t}={sinx}\:\:{dt}={cosxdx} \\ $$$$\int\frac{{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+{t}+\mathrm{1}−{t}}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}−{t}\right)}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\mathrm{1}−{t}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\mathrm{1}+{t}} \\ $$$$ \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{1}−{t}\mid+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\mathrm{1}+{t}\mid \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\mathrm{1}+{sinx}}{\mathrm{1}−{sinx}}\mid+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\left(\mathrm{1}+{sinx}\right)^{\mathrm{2}} }{\mathrm{1}−{sin}^{\mathrm{2}} {x}}\mid+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\left(\mathrm{1}+{sinx}\right)^{\mathrm{2}} }{{cos}^{\mathrm{2}} {x}}\mid+{c} \\ $$$$={ln}\mid\frac{\mathrm{1}}{{cosx}}+\frac{{sinx}}{{cosx}}\mid+{c} \\ $$$$={ln}\mid{secx}+{tanx}\mid+{c} \\ $$$$ \\ $$$$ \\ $$

Answered by malwaan last updated on 25/Jun/18

∫((secx(secx+tanx))/(secx+tanx)) =∫((sec^2 x+secx.tanx)/(tanx+secx)) =ln∣tanx+secx∣+c

$$\int\frac{\mathrm{secx}\left(\mathrm{secx}+\mathrm{tanx}\right)}{\mathrm{secx}+\mathrm{tanx}} \\ $$$$=\int\frac{\mathrm{sec}^{\mathrm{2}} \mathrm{x}+\mathrm{secx}.\mathrm{tanx}}{\mathrm{tanx}+\mathrm{secx}} \\ $$$$=\mathrm{ln}\mid\mathrm{tanx}+\mathrm{secx}\mid+\mathrm{c} \\ $$

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