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Question Number 93958 by mashallah last updated on 16/May/20

∫((sinx−cosx)/(sinx+cosx))dx=

$$\int\frac{\mathrm{sinx}−\mathrm{cosx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}= \\ $$

Commented by mathmax by abdo last updated on 17/May/20

I =∫ ((sinx−cosx)/(sinx +cosx))dx ⇒I =∫((tanx−1)/(tanx +1))dx =_(tanx =t) =∫((t−1)/(t+1))×(dt/(1+t^2 )) =∫ ((t−1)/((t+1)(t^2 +1)))dt let decompose F(t) =((t−1)/((t+1)(t^2 +1))) ⇒F(t) =(a/(t+1)) +((bt+c)/(t^2 +1)) a =−1 lim_(t→+∞) tF(t) =0 =a+b ⇒b =1 F(o) =−1 =a +c ⇒c =0 ⇒F(t) =−(1/(t+1)) +(t/(t^2 +1)) ⇒ I =−ln∣t+1∣+(1/2)ln(1+t^2 ) +C =(1/2)ln(1+tan^2 x)−ln∣1+tanx∣ +C

$${I}\:=\int\:\frac{{sinx}−{cosx}}{{sinx}\:+{cosx}}{dx}\:\Rightarrow{I}\:=\int\frac{{tanx}−\mathrm{1}}{{tanx}\:+\mathrm{1}}{dx}\:=_{{tanx}\:={t}} \\ $$$$=\int\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}×\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int\:\:\:\:\frac{{t}−\mathrm{1}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\frac{{t}−\mathrm{1}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow{F}\left({t}\right)\:=\frac{{a}}{{t}+\mathrm{1}}\:+\frac{{bt}+{c}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:=−\mathrm{1}\:\:\:{lim}_{{t}\rightarrow+\infty} {tF}\left({t}\right)\:=\mathrm{0}\:={a}+{b}\:\Rightarrow{b}\:=\mathrm{1}\: \\ $$$${F}\left({o}\right)\:=−\mathrm{1}\:={a}\:+{c}\:\Rightarrow{c}\:=\mathrm{0}\:\Rightarrow{F}\left({t}\right)\:=−\frac{\mathrm{1}}{{t}+\mathrm{1}}\:+\frac{{t}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${I}\:=−{ln}\mid{t}+\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)−{ln}\mid\mathrm{1}+{tanx}\mid\:+{C} \\ $$

Answered by john santu last updated on 16/May/20

(((sin x−cos x)^2 )/(−cos 2x)) = ((1−sin 2x)/(−cos 2x)) = tan 2x−sec 2x

$$\frac{\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }{−\mathrm{cos}\:\mathrm{2x}}\:=\:\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{2x}}{−\mathrm{cos}\:\mathrm{2x}} \\ $$$$=\:\mathrm{tan}\:\mathrm{2x}−\mathrm{sec}\:\mathrm{2x}\: \\ $$$$ \\ $$

Answered by Kunal12588 last updated on 16/May/20

sin x+cos x=t ⇒−(sin x−cos x)dx=dt I=∫−(dt/t)=−ln∣sin x + cos x∣

$${sin}\:{x}+{cos}\:{x}={t} \\ $$$$\Rightarrow−\left({sin}\:{x}−{cos}\:{x}\right){dx}={dt} \\ $$$${I}=\int−\frac{{dt}}{{t}}=−{ln}\mid{sin}\:{x}\:+\:{cos}\:{x}\mid \\ $$

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