1-sinx-cosx-dx-

Question Number 114072 by Lordose last updated on 17/Sep/20

$$\int\frac{\mathrm{1}}{\boldsymbol{\mathrm{sinx}}\:+\:\boldsymbol{\mathrm{cosx}}}\boldsymbol{\mathrm{dx}} \\ $$

Answered by Olaf last updated on 17/Sep/20

x = (π/4)−u sinx+cosx = (sin(π/4)cosu−sinucos(π/4))+(cos(π/4)cosu+sin(π/4)sinu) = ((√2)/2)(cosu−sinu)+((√2)/2)(cosu+sinu) = (√2)cosu −(1/( (√2)))∫(du/(cosu)) = −(1/( (√2)))∫(du/((1−t^2 )/(1+t^2 ))) with t = tan(u/2) dt = (1/2)(1+tan^2 (u/2))du du = ((2dt)/(1+t^2 )) −(1/( (√2)))∫((1+t^2 )/(1−t^2 )).((2dt)/(1+t^2 )) = −(√2)∫(dt/(1−t^2 )) = −(1/( (√2)))∫((1/(1−t))+(1/(1+t)))dt = (1/( (√2)))ln∣((1−t)/(1+t))∣ = (1/( (√2)))ln∣((1−tan(u/2))/(1+tan(u/2)))∣ (1/( (√2)))ln∣((1−tan((π/8)−(x/2)))/(1+tan((π/8)−(x/2))))∣...

$${x}\:=\:\frac{\pi}{\mathrm{4}}−{u} \\ $$$$\mathrm{sin}{x}+\mathrm{cos}{x}\:=\: \\ $$$$\left(\mathrm{sin}\frac{\pi}{\mathrm{4}}\mathrm{cos}{u}−\mathrm{sin}{u}\mathrm{cos}\frac{\pi}{\mathrm{4}}\right)+\left(\mathrm{cos}\frac{\pi}{\mathrm{4}}\mathrm{cos}{u}+\mathrm{sin}\frac{\pi}{\mathrm{4}}\mathrm{sin}{u}\right) \\ $$$$=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{cos}{u}−\mathrm{sin}{u}\right)+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{cos}{u}+\mathrm{sin}{u}\right) \\ $$$$=\:\sqrt{\mathrm{2}}\mathrm{cos}{u} \\ $$$$−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{du}}{\mathrm{cos}{u}}\:=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{du}}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$$\mathrm{with}\:{t}\:=\:\mathrm{tan}\frac{{u}}{\mathrm{2}} \\ $$$${dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \frac{{u}}{\mathrm{2}}\right){du} \\ $$$${du}\:=\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }.\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:−\sqrt{\mathrm{2}}\int\frac{{dt}}{\mathrm{1}−{t}^{\mathrm{2}} }\:=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}+\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\mid\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\mid\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\mid\frac{\mathrm{1}−\mathrm{tan}\frac{{u}}{\mathrm{2}}}{\mathrm{1}+\mathrm{tan}\frac{{u}}{\mathrm{2}}}\mid \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{ln}\mid\frac{\mathrm{1}−\mathrm{tan}\left(\frac{\pi}{\mathrm{8}}−\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}+\mathrm{tan}\left(\frac{\pi}{\mathrm{8}}−\frac{{x}}{\mathrm{2}}\right)}\mid… \\ $$

Answered by MJS_new last updated on 17/Sep/20

∫(dx/(sin x +cos x))= [t=x+(π/4) → dx=dt] =((√2)/2)∫csc t dt=((√2)/2)ln ∣tan (t/2)∣= =((√2)/2)ln ∣tan ((x/2)+(π/8))∣= =((√2)/2)ln ∣(((√2)+2sin x)/( (√2)+2cos x))∣ +C

$$\int\frac{{dx}}{\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\frac{\pi}{\mathrm{4}}\:\rightarrow\:{dx}={dt}\right] \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\mathrm{csc}\:{t}\:{dt}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\mid= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)\mid= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{\sqrt{\mathrm{2}}+\mathrm{2sin}\:{x}}{\:\sqrt{\mathrm{2}}+\mathrm{2cos}\:{x}}\mid\:+{C} \\ $$

Answered by Dwaipayan Shikari last updated on 17/Sep/20

∫(1/(sinx+cosx))dx =(√2)∫(1/(cos((π/4)−x)))dx =(√2) ∫sec((π/4)−x)dx =−(√2) ∫secudu=−(√2) log(secu+tanu) =−(√2) log(sec((π/4)−x)+tan((π/4)−x))

$$\int\frac{\mathrm{1}}{{sinx}+{cosx}}{dx} \\ $$$$=\sqrt{\mathrm{2}}\int\frac{\mathrm{1}}{{cos}\left(\frac{\pi}{\mathrm{4}}−{x}\right)}{dx} \\ $$$$=\sqrt{\mathrm{2}}\:\int{sec}\left(\frac{\pi}{\mathrm{4}}−{x}\right){dx} \\ $$$$=−\sqrt{\mathrm{2}}\:\int{secudu}=−\sqrt{\mathrm{2}}\:{log}\left({secu}+{tanu}\right) \\ $$$$=−\sqrt{\mathrm{2}}\:\:{log}\left({sec}\left(\frac{\pi}{\mathrm{4}}−{x}\right)+{tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right) \\ $$

Commented by MJS_new last updated on 17/Sep/20

$$\mathrm{no}. \\ $$

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