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Question Number 64333 by Chi Mes Try last updated on 16/Jul/19

∫_( 0) ^(π/2) (1/(1+tan x)) dx =

$$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}\:{x}}\:{dx}\:= \\ $$

Commented by mathmax by abdo last updated on 16/Jul/19

let I =∫_0 ^(π/2) (dx/(1+tanx)) changement tanx =t give I =∫_0 ^(+∞) (dt/((1+t^2 )(1+t))) let decompose F(t)=(1/((t+1)(t^2 +1))) F(t)=(a/(t+1)) +((bt+c)/(t^2 +1)) a=lim_(t→−1) (t+1)F(t) =(1/2) lim_(t→+∞) tF(t) =0 =a+b ⇒b=−(1/2) ⇒F(t) =(1/(2(t+1))) +((−(1/2)t +c)/(t^2 +1)) F(0)=1 =(1/2) +c ⇒c=(1/2) ⇒F(t) =(1/(2(t+1))) −(1/2) ((t−1)/(t^2 +1)) ⇒ I =∫_0 ^∞ ((1/(2(t+1)))−(1/4)((2t)/(t^2 +1)))dt +(1/2)∫_0 ^∞ (dt/(t^2 +1)) =(1/2)[ln∣t+1∣−(1/2)ln(t^2 +1)]_0 ^(+∞) +(π/4) =(1/2)[ln∣((t+1)/(√(t^2 +1)))∣]_0 ^(+∞) +(π/4) =0 +(π/4) ⇒ I =(π/4) .

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\mathrm{1}+{tanx}}\:{changement}\:{tanx}\:={t}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}\right)}\:\:{let}\:{decompose}\:{F}\left({t}\right)=\frac{\mathrm{1}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${F}\left({t}\right)=\frac{{a}}{{t}+\mathrm{1}}\:+\frac{{bt}+{c}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}={lim}_{{t}\rightarrow−\mathrm{1}} \left({t}+\mathrm{1}\right){F}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${lim}_{{t}\rightarrow+\infty} \:{tF}\left({t}\right)\:=\mathrm{0}\:={a}+{b}\:\Rightarrow{b}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{F}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}\:+\frac{−\frac{\mathrm{1}}{\mathrm{2}}{t}\:+{c}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{2}}\:+{c}\:\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{F}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{4}}\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\right){dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid{t}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{+\infty} \:+\frac{\pi}{\mathrm{4}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\frac{{t}+\mathrm{1}}{\sqrt{{t}^{\mathrm{2}} \:+\mathrm{1}}}\mid\right]_{\mathrm{0}} ^{+\infty} \:+\frac{\pi}{\mathrm{4}}\:=\mathrm{0}\:+\frac{\pi}{\mathrm{4}}\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{4}}\:. \\ $$$$ \\ $$

Commented by Tony Lin last updated on 17/Jul/19

∫_0 ^(π/2) (1/(1+tanx))dx =∫_0 ^(π/2) ((cosx)/(cosx+sinx))dx =∫_0 ^(π/2) ((cos(cosx−sinx))/((cosx+sinx)(cosx−sinx)))dx =∫_0 ^(π/2) ((cos^2 x−cosxsinx)/(cos^2 x−sin^2 x))dx =(1/2)∫_0 ^(π/2) ((cos2x+1−sin2x)/(cos2x))dx =(1/2)∫_0 ^(π/2) dx−(1/2)∫_0 ^(π/2) tan2xdx+(1/2)∫_0 ^(π/2) sec2xdx =(π/4)

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+{tanx}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cosx}}{{cosx}+{sinx}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({cosx}−{sinx}\right)}{\left({cosx}+{sinx}\right)\left({cosx}−{sinx}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{2}} {x}−{cosxsinx}}{{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\mathrm{2}{x}+\mathrm{1}−{sin}\mathrm{2}{x}}{{cos}\mathrm{2}{x}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {tan}\mathrm{2}{xdx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sec}\mathrm{2}{xdx} \\ $$$$=\frac{\pi}{\mathrm{4}} \\ $$

Answered by Rio Michael last updated on 16/Jul/19

begin with ∫(1/(1+tanx))dx I = ∫(1/(1+tanx))dx use the substitution u = tanx I = ∫(1/((1+u^2 )(1+u)))du I = (1/2)∫((1/(1+u^2 )) + (1/(1+u)))du I = (1/2)∫((1/(1+u^2 ))−(1/2) ((2u)/(1+u^2 )) + (1/(1+u)))du I = (1/2)(arctanu−(1/2)ln(1+u^2 )+ln(1+u)) I = ⌈ (1/2)(x−ln(secx) + ln(1+tanx))⌉_0 ^(π/2) I = (1/2)((π/2)+ln(sin(π/2)+cos(π/2))) = (π/4)

$${begin}\:{with} \\ $$$$\int\frac{\mathrm{1}}{\mathrm{1}+{tanx}}{dx} \\ $$$${I}\:=\:\int\frac{\mathrm{1}}{\mathrm{1}+{tanx}}{dx}\:\:\:\:\:\:{use}\:{the}\:{substitution}\:{u}\:=\:{tanx} \\ $$$${I}\:=\:\int\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{u}\right)}{du} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}+{u}}\right){du} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}+{u}}\right){du} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({arctanu}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)+{ln}\left(\mathrm{1}+{u}\right)\right) \\ $$$${I}\:=\:\lceil\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}−{ln}\left({secx}\right)\:+\:{ln}\left(\mathrm{1}+{tanx}\right)\right)\rceil_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}+{ln}\left({sin}\frac{\pi}{\mathrm{2}}+{cos}\frac{\pi}{\mathrm{2}}\right)\right)\:=\:\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$

Answered by Tanmay chaudhury last updated on 17/Jul/19

I=∫_0 ^(π/2) ((cosx)/(cosx+sinx))dx =∫_0 ^(π/2) ((cos((π/2)−x))/(cos((π/2)−x)+sin((π/2)−x)))dx =∫_0 ^(π/2) ((sinx)/(sinx+cox))dx 2I=∫_0 ^(π/2) ((cosx)/(sinx+cosx))+((sinx)/(cosx+sinx))dx 2I=∫_0 ^(π/2) dx I=(π/4)

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cosx}}{{cosx}+{sinx}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)+{sin}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sinx}}{{sinx}+{cox}}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cosx}}{{sinx}+{cosx}}+\frac{{sinx}}{{cosx}+{sinx}}{dx} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\ $$$${I}=\frac{\pi}{\mathrm{4}} \\ $$

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