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Question Number 55493 by Kunal12588 last updated on 25/Feb/19

Find intgral of tan x by parts ∫tan x dx pls help me (i am a newbie in calculus)

$${Find}\:{intgral}\:{of}\:\mathrm{tan}\:{x}\:{by}\:{parts} \\ $$$$\int{tan}\:{x}\:{dx}\:\:\:{pls}\:{help}\:{me}\:\:\left({i}\:{am}\:{a}\:{newbie}\:{in}\:{calculus}\right) \\ $$

Commented by Kunal12588 last updated on 25/Feb/19

i am getting something like this ∫((1/(cosx)))×(sin x dx) taking u=(1/(cosx)) ⇒du= tan x sec x dx taking dv=sinx dx ⇒v=−cosx ∫u dv = uv − ∫v du ∫tan x dx = (1/(cos x))×(−cos x) −∫−cos x × tan x sec x dx ∫tan x dx = −1+∫tan x dx but in the book :− ∫tan x dx = −ln cos x

$${i}\:{am}\:{getting}\:{something}\:{like}\:{this} \\ $$$$\int\left(\frac{\mathrm{1}}{{cosx}}\right)×\left({sin}\:{x}\:{dx}\right) \\ $$$${taking}\:{u}=\frac{\mathrm{1}}{{cosx}} \\ $$$$\Rightarrow{du}=\:{tan}\:{x}\:{sec}\:{x}\:{dx} \\ $$$${taking}\:{dv}={sinx}\:{dx} \\ $$$$\Rightarrow{v}=−{cosx} \\ $$$$\int{u}\:{dv}\:=\:{uv}\:−\:\int{v}\:{du} \\ $$$$\int{tan}\:{x}\:{dx}\:=\:\frac{\mathrm{1}}{{cos}\:{x}}×\left(−{cos}\:{x}\right)\:−\int−{cos}\:{x}\:×\:{tan}\:{x}\:{sec}\:{x}\:{dx} \\ $$$$\int{tan}\:{x}\:{dx}\:=\:−\mathrm{1}+\int{tan}\:{x}\:{dx} \\ $$$${but}\:{in}\:{the}\:{book}\::− \\ $$$$\int{tan}\:{x}\:{dx}\:=\:−{ln}\:{cos}\:{x} \\ $$

Commented by maxmathsup by imad last updated on 25/Feb/19

let I =∫ tanx dx ⇒I =∫ ((sinx)/(cosx))dx =_(tan((x/2))=t) ∫ (((2t)/(1+t^2 ))/((1−t^2 )/(1+t^2 ))) ((2dt)/(1+t^2 )) = ∫ ((4t)/((1+t^2 )(1−t^2 )))dt =∫ 2t{(1/(1+t^2 )) +(1/(1−t^2 ))}dt =∫ ((2t)/(t^2 +1))dt −∫ ((2t)/(t^2 −1))dt =ln∣t^2 +1∣−ln∣t^2 −1∣ +c =ln∣((1+tan^2 ((x/2)))/(1−tan^2 ((x/2))))∣ +c =ln∣((1+((sin^2 ((x/2)))/(cos^2 ((x/2)))))/(1−((sin^2 ((x/2)))/(cos^2 ((x/2))))))∣ +c =ln∣ (1/(cos(x)))∣ +c =ln∣cosx∣ +C .

$${let}\:{I}\:=\int\:{tanx}\:{dx}\:\Rightarrow{I}\:=\int\:\frac{{sinx}}{{cosx}}{dx}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int\:\:\frac{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\frac{\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{dt}\:=\int\:\:\mathrm{2}{t}\left\{\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{1}−{t}^{\mathrm{2}} }\right\}{dt} \\ $$$$=\int\:\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}\:−\int\:\:\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\:\:={ln}\mid{t}^{\mathrm{2}} +\mathrm{1}\mid−{ln}\mid{t}^{\mathrm{2}} −\mathrm{1}\mid\:+{c} \\ $$$$={ln}\mid\frac{\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}\mid\:+{c}\:={ln}\mid\frac{\mathrm{1}+\frac{{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}}{\mathrm{1}−\frac{{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}}\mid\:+{c} \\ $$$$={ln}\mid\:\frac{\mathrm{1}}{{cos}\left({x}\right)}\mid\:+{c}\:={ln}\mid{cosx}\mid\:+{C}\:. \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Feb/19

what is the question ∫tanxdx→by parts or by partial fraction...pls recheck the question..

$${what}\:{is}\:{the}\:{question}\:\int{tanxdx}\rightarrow{by}\:{parts} \\ $$$${or}\:{by}\:\boldsymbol{{partial}}\:\boldsymbol{{fraction}}...\boldsymbol{{pls}}\:\boldsymbol{{recheck}}\:\boldsymbol{{the}}\:\boldsymbol{{question}}.. \\ $$

Commented by maxmathsup by imad last updated on 25/Feb/19

I =−ln∣cosx∣ +C .

$${I}\:=−{ln}\mid{cosx}\mid\:+{C}\:\:. \\ $$

Commented by maxmathsup by imad last updated on 25/Feb/19

i have used changement method not integration by parts...

$$\:{i}\:{have}\:{used}\:{changement}\:{method}\:{not}\:{integration}\:{by}\:{parts}... \\ $$

Commented by Kunal12588 last updated on 26/Feb/19

sir it was not a question I was trying by my own. I want to know where is the mistake

$${sir}\:{it}\:{was}\:{not}\:{a}\:{question}\:{I}\:{was}\:{trying}\:{by}\:{my} \\ $$$${own}.\:{I}\:{want}\:{to}\:{know}\:{where}\:{is}\:{the}\:{mistake} \\ $$

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