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Question Number 35059 by math khazana by abdo last updated on 14/May/18
find ∫_0 ^π (dx/(cosx +sinx))
$${find}\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\frac{{dx}}{{cosx}\:+{sinx}} \\ $$
Commented by math khazana by abdo last updated on 15/May/18
let put I = ∫_0 ^π (dx/(cosx +sinx)) changement tan((x/2))=t give x=2arctant and I = ∫_0 ^(+∞) (1/(((1−t^2 )/(1+t^2 )) +((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) I = ∫_0 ^(+∞) ((2dt)/(1−t^2 +2t)) = −2 ∫_0 ^∞ (dt/(t^2 −2t −1)) =−2 ∫_0 ^∞ (dt/((t−1)^2 −2)) =−2 ∫_0 ^∞ (dt/((t−3)(t+1))) =(1/2) ∫_0 ^∞ { (1/(t+1)) −(1/(t−3))}dt =(1/2) [ln∣((t+1)/(t−3))∣]_0 ^(+∞) =(1/2)( −ln((1/3))) I =(1/2)ln(3).
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{{cosx}\:+{sinx}}\:\:{changement} \\ $$$${tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give}\:\:{x}=\mathrm{2}{arctant}\:{and} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} \:+\mathrm{2}{t}}\:=\:−\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:−\mathrm{2}{t}\:−\mathrm{1}} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{2}} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{dt}}{\left({t}−\mathrm{3}\right)\left({t}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \left\{\:\frac{\mathrm{1}}{{t}+\mathrm{1}}\:−\frac{\mathrm{1}}{{t}−\mathrm{3}}\right\}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\left[{ln}\mid\frac{{t}+\mathrm{1}}{{t}−\mathrm{3}}\mid\right]_{\mathrm{0}} ^{+\infty} \:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\:−{ln}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\right) \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{3}\right). \\ $$
Commented by prof Abdo imad last updated on 16/May/18
error in the final lines I =−2 ∫_0 ^∞ (dt/((t−1)^2 −2)) =−2 ∫_0 ^∞ (dt/((t−1−_ (√2))(t−1 +(√2)))) =((−2)/(2(√2)))∫_0 ^∞ { (1/(t−1−(√2))) −(1/(t−1+(√2)))}dt =−(1/( (√2)))[ln∣((t−1−(√2))/(t−1+(√2)))]]_0 ^(+∞) =((−1)/( (√2)))( −ln(((1+(√2))/( (√2) −1)))) I = (1/( (√2)))ln(((1+(√2))/(−1+(√2))))
$${error}\:{in}\:{the}\:{final}\:{lines} \\ $$$${I}\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{dt}}{\left({t}−\mathrm{1}−_{} \sqrt{\mathrm{2}}\right)\left({t}−\mathrm{1}\:+\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{−\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\left\{\:\:\frac{\mathrm{1}}{{t}−\mathrm{1}−\sqrt{\mathrm{2}}}\:−\frac{\mathrm{1}}{{t}−\mathrm{1}+\sqrt{\mathrm{2}}}\right\}{dt} \\ $$$$\left.=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[{ln}\mid\frac{{t}−\mathrm{1}−\sqrt{\mathrm{2}}}{{t}−\mathrm{1}+\sqrt{\mathrm{2}}}\right]\right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\:−{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}\:−\mathrm{1}}\right)\right)\: \\ $$$${I}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{−\mathrm{1}+\sqrt{\mathrm{2}}}\right) \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/May/18
I=∫_0 ^Π (dx/(cos(Π−x)+sin(Π−x))) 2I=∫_0 ^Π (1/(cosx+sinx))+(1/(−cosx+sinx)) dx =∫_0 ^Π ((2sinx)/(sin^2 x−cos^2 x)) dx I=(1/2)∫_0 ^Π ((2sinx)/(1−2cos^2 x))dx =∫_0 ^Π (((−sinx))/(2cos^2 x−1))dx =(1/2)∫_0 ^Π ((d(cosx))/(cos^2 x−((1/( (√2))))^2 )) =(1/2)×((√2)/2)∣ln((cosx−(1/( (√2))))/(cosx+(1/( (√2)))))∣_0 ^Π
$${I}=\int_{\mathrm{0}} ^{\Pi} \frac{{dx}}{{cos}\left(\Pi−{x}\right)+{sin}\left(\Pi−{x}\right)} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\Pi} \frac{\mathrm{1}}{{cosx}+{sinx}}+\frac{\mathrm{1}}{−{cosx}+{sinx}}\:{dx} \\ $$$$=\int_{\mathrm{0}} ^{\Pi} \frac{\mathrm{2}{sinx}}{{sin}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}}\:{dx} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\Pi} \frac{\mathrm{2}{sinx}}{\mathrm{1}−\mathrm{2}{cos}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\Pi} \frac{\left(−{sinx}\right)}{\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\Pi} \frac{{d}\left({cosx}\right)}{{cos}^{\mathrm{2}} {x}−\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mid{ln}\frac{{cosx}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}{{cosx}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\mid_{\mathrm{0}} ^{\Pi} \\ $$
Commented by NECx last updated on 15/May/18
please how did you change ∫_0 ^π (1/(cosx+sinx))dx to ∫_0 ^π (1/(cos(π−x)+sin(π−x)))dx???
$${please}\:{how}\:{did}\:{you}\:{change} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{{cosx}+{sinx}}{dx}\:{to} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{{cos}\left(\pi−{x}\right)+{sin}\left(\pi−{x}\right)}{dx}??? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 15/May/18
∫_0 ^a f(x)dx=∫_0 ^a f(a−x)dx
$$\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{a}} {f}\left({a}−{x}\right){dx} \\ $$
Commented by NECx last updated on 15/May/18
oh... Thanks
$${oh}…\:{Thanks} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/May/18
∫(dx/(cosx+sinx)) t=tan(x/2) dt=(1/2)sec^2 (x/2) ∫(dx/(((1−tan^2 (x/2))/(1+tan^2 (x/2)))+((2tan(x/2))/(1+tan^2 (x/2))))) ∫((sec^2 (x/2))/(1−tan^2 (x/2)+2tan(x/2))) ∫((2dt)/(1−t^2 +2t)) −2∫(dt/(t^2 −2t+1−2)) −2∫(dt/((t−1)^2 −((√2))^2 )) now use formula and put limit or method ∫_0 ^Π (dx/(cosx+sinx)) =∫_0 ^Π (((1/( (√2)))dx)/((1/( (√2)))cosx+(1/( (√2)))sinx)) =(1/( (√2)))∫_0 ^Π (dx/(sin(Π/4+x))) now useformula and put limit
$$\int\frac{{dx}}{{cosx}+{sinx}} \\ $$$${t}={tan}\frac{{x}}{\mathrm{2}}\:\:{dt}=\frac{\mathrm{1}}{\mathrm{2}}{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}} \\ $$$$\int\frac{{dx}}{\frac{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}+\frac{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}} \\ $$$$\int\frac{{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}} \\ $$$$\int\frac{\mathrm{2}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} +\mathrm{2}{t}} \\ $$$$−\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}−\mathrm{2}} \\ $$$$−\mathrm{2}\int\frac{{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${now}\:{use}\:{formula}\:{and}\:{put}\:{limit} \\ $$$$ \\ $$$${or}\:{method} \\ $$$$\int_{\mathrm{0}} ^{\Pi} \frac{{dx}}{{cosx}+{sinx}} \\ $$$$=\int_{\mathrm{0}} ^{\Pi} \frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{dx}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{cosx}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sinx}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\Pi} \frac{{dx}}{{sin}\left(\Pi/\mathrm{4}+{x}\right)} \\ $$$$ \\ $$$$ \\ $$$${now}\:{useformula}\:{and}\:{put}\:{limit} \\ $$$$ \\ $$

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