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Question Number 37237 by abdo.msup.com last updated on 11/Jun/18
calculate ∫_0 ^(π/2) ((cosθ.sinθ)/(cosθ +sinθ)) dθ .
$${calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{cos}\theta.{sin}\theta}{{cos}\theta\:+{sin}\theta}\:{d}\theta\:. \\ $$
Commented by abdo.msup.com last updated on 27/Jul/18
changement tan((θ/2))=t give I = ∫_0 ^1 (((2t(1−t^2 ))/((1+t^2 )^2 ))/(((1−t^2 )/(1+t^2 )) +((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫_0 ^1 ((2t(1−t^2 ))/((1+t^2 )^2 (1−t^2 +2t)))dt = ∫_0 ^1 ((2t(t^2 −1))/((1+t^2 )^2 (t−1)^2 ))dt =∫_0 ^1 ((2t(t+1))/((t^2 +1)^2 (t−1)))dt =∫_0 ^1 ((2t^2 +2t)/((t−1)(t^2 +1)^2 ))dt let decompose F(t)=((2t^2 +2t)/((t−1)(t^2 +1)^2 )) F(t)=(a/(t−1)) +((bt+c)/(t^2 +1)) +((dt +e)/((t^2 +1)^2 )) ....be continued...
$${changement}\:{tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} +\mathrm{2}{t}\right)}{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{t}\left({t}+\mathrm{1}\right)}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \left({t}−\mathrm{1}\right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}{t}}{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:{let}\:{decompose} \\ $$$${F}\left({t}\right)=\frac{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}{t}}{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${F}\left({t}\right)=\frac{{a}}{{t}−\mathrm{1}}\:+\frac{{bt}+{c}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{dt}\:+{e}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$….{be}\:{continued}… \\ $$
Answered by math1967 last updated on 11/Jun/18
(1/2)∫((2sinθcosθ)/(cosθ+sinθ))dθ (1/2)∫(((sinθ+cosθ)^2 )/(cosθ+sinθ))dθ −(1/2)∫(dθ/(cosθ+sinθ)) −(1/2)cosθ+(1/2)sinθ−(1/(2(√2)))∫(dθ/(sin(π/4)cosθ+cos(π/4)sinθ)) (1/2)(sinθ−cosθ)−(1/(2(√2)))∫cosec((π/4)+θ)dθ (1/2)(sinθ−cosθ)−(1/(2(√2)))ln[cosec((π/4)+θ)−cot((π/4)+θ)] ∴∫_0 ^(π/2) ((sinθcosθdθ)/(sinθ+cosθ))={(1/2)×1−(1/(2(√2)))ln[(√2) +1]} −{−(1/2)−(1/(2(√2)))ln[[(√2) −1]}=1+(1/(2(√2)))ln(((√2) −1)/( (√2) +1))
$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{sin}\theta{cos}\theta}{{cos}\theta+{sin}\theta}{d}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({sin}\theta+{cos}\theta\right)^{\mathrm{2}} }{{cos}\theta+{sin}\theta}{d}\theta\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\theta}{{cos}\theta+{sin}\theta} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}{cos}\theta+\frac{\mathrm{1}}{\mathrm{2}}{sin}\theta−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\frac{{d}\theta}{{sin}\frac{\pi}{\mathrm{4}}{cos}\theta+{cos}\frac{\pi}{\mathrm{4}}{sin}\theta} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({sin}\theta−{cos}\theta\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int{cosec}\left(\frac{\pi}{\mathrm{4}}+\theta\right){d}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({sin}\theta−{cos}\theta\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left[{cosec}\left(\frac{\pi}{\mathrm{4}}+\theta\right)−{cot}\left(\frac{\pi}{\mathrm{4}}+\theta\right)\right] \\ $$$$\therefore\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{sin}\theta{cos}\theta{d}\theta}{{sin}\theta+{cos}\theta}=\left\{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left[\sqrt{\mathrm{2}}\:+\mathrm{1}\right]\right\} \\ $$$$−\left\{−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left[\left[\sqrt{\mathrm{2}}\:−\mathrm{1}\right]\right\}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\frac{\sqrt{\mathrm{2}}\:−\mathrm{1}}{\:\sqrt{\mathrm{2}}\:+\mathrm{1}}\right. \\ $$
Commented by math1967 last updated on 11/Jun/18
Pls.check
$${Pls}.{check} \\ $$

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