secxdx-

Question Number 43157 by MASANJA J last updated on 07/Sep/18

$$\int{secxdx} \\ $$

Commented by maxmathsup by imad last updated on 07/Sep/18

let J = ∫ (dx/(cosx)) changement tan((x/2))=u give J = ∫ (1/((1−u^2 )/(1+u^2 ))) ((2du)/(1+u^2 )) = ∫ ((2du)/(1−u^2 )) = ∫ ((1/(1+u)) +(1/(1−u)))du =ln∣((1+u)/(1−u))∣ +c =ln∣ ((1+tan((x/2)))/(1−tan((x/2))))∣ +c =ln∣tan((x/2) +(π/4))∣ +c .

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

Commented by $@ty@m last updated on 08/Sep/18

Pl. see my comment on Q. No. 40717 for two step solution.

$${Pl}.\:{see}\:{my}\:{comment}\:{on}\:{Q}.\:{No}. \\ $$$$\mathrm{40717} \\ $$$${for}\:{two}\:{step}\:{solution}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Sep/18

∫(dx/(cosx)) ∫((cosx dx)/(1−sin^2 x)) t=sinx dt=cosx dx ∫(dt/(1−t^2 )) (1/2)∫((1+t+1−t)/((1+t)(1−t)))dt (1/2)[∫(dt/(1−t))+∫(dt/(1+t))] k=1−t dk=−dt (1/2)[∫((−dk)/k)+∫(dt/(1+t))] (1/2)[ln(1+t)−lnk] (1/2)ln∣((1+t)/(1−t))∣+c k=1−t (1/2)ln∣(((1+t)^2 )/(1−t^2 ))∣+c (1/2)ln∣(((1+sinx)^2 )/(1−sin^2 x))∣+c (1/2)ln∣(((1+sinx)/(cosx)))^2 ∣+c ln∣(1/(cosx))+((sinx)/(cosx))∣+c ln∣secx+tanx∣+c

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

Facebook Tweet Pin

Leave a Reply Cancel reply