Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 163829 by milandou last updated on 11/Jan/22

∫(1/(cos x))

$$\int\frac{\mathrm{1}}{\mathrm{cos}\:{x}} \\ $$

Commented by MJS_new last updated on 11/Jan/22

∫x+y=?  maybe use proper syntax?

$$\int{x}+{y}=? \\ $$$$\mathrm{maybe}\:\mathrm{use}\:\mathrm{proper}\:\mathrm{syntax}? \\ $$

Answered by Ar Brandon last updated on 11/Jan/22

∫(1/(cosx))dx=ln(secx+tanx)dx  Let x=(π/2)−2t⇒dx=−2dt  ∫(dx/(cosx))=−2∫(1/(sin2t))dt=−2∫((cos^2 t+sin^2 t)/(2sintcost))dt  =−∫(((cost)/(sint))+((sint)/(cost)))dt=ln(cost)−ln(sint)+C  =ln(((cost)/(sint)))+C=ln∣cot((π/4)−(x/2))∣+C

$$\int\frac{\mathrm{1}}{\mathrm{cos}{x}}{dx}=\mathrm{ln}\left(\mathrm{sec}{x}+\mathrm{tan}{x}\right){dx} \\ $$$$\mathrm{Let}\:{x}=\frac{\pi}{\mathrm{2}}−\mathrm{2}{t}\Rightarrow{dx}=−\mathrm{2}{dt} \\ $$$$\int\frac{{dx}}{\mathrm{cos}{x}}=−\mathrm{2}\int\frac{\mathrm{1}}{\mathrm{sin2}{t}}{dt}=−\mathrm{2}\int\frac{\mathrm{cos}^{\mathrm{2}} {t}+\mathrm{sin}^{\mathrm{2}} {t}}{\mathrm{2sin}{t}\mathrm{cos}{t}}{dt} \\ $$$$=−\int\left(\frac{\mathrm{cos}{t}}{\mathrm{sin}{t}}+\frac{\mathrm{sin}{t}}{\mathrm{cos}{t}}\right){dt}=\mathrm{ln}\left(\mathrm{cos}{t}\right)−\mathrm{ln}\left(\mathrm{sin}{t}\right)+{C} \\ $$$$=\mathrm{ln}\left(\frac{\mathrm{cos}{t}}{\mathrm{sin}{t}}\right)+{C}=\mathrm{ln}\mid\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}−\frac{{x}}{\mathrm{2}}\right)\mid+{C} \\ $$

Answered by Ar Brandon last updated on 11/Jan/22

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

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

Answered by stelor last updated on 12/Jan/22

I=∫(1/(cos x))dx  u=sinx  du=cosxdx and (dx/(cosx)) = (du/(1−u^2 ))  I=∫(du/(1−u^2 ))  I = argth(u) + c  I = (1/2)ln(((1+u)/(1−u))) + c  I=(1/2)ln(((1+ sinx)/(1−sinx))) + c

$${I}=\int\frac{\mathrm{1}}{\mathrm{cos}\:{x}}{dx} \\ $$$${u}={sinx} \\ $$$${du}={cosxdx}\:{and}\:\frac{{dx}}{{cosx}}\:=\:\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$${I}=\int\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$${I}\:=\:{argth}\left({u}\right)\:+\:{c} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\right)\:+\:{c} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+\:{sinx}}{\mathrm{1}−{sinx}}\right)\:+\:{c} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com