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Question Number 129180 by mnjuly1970 last updated on 13/Jan/21

... nice calculus... calculate:: φ=^(???) ∫_0 ^( (π/4)) (dx/((sin(x)+cos(x)+(√2) )^2 ))

$$\:\:\:\:\:\:\:\:\:\:\:\:...\:{nice}\:\:\:{calculus}... \\ $$$$\:\:\:{calculate}:: \\ $$$$\:\:\:\:\:\:\phi\overset{???} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\left({sin}\left({x}\right)+{cos}\left({x}\right)+\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} } \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 13/Jan/21

Integral doesn′t Converge

$${Integral}\:{doesn}'{t}\:{Converge} \\ $$

Commented by mnjuly1970 last updated on 13/Jan/21

thank you for your attantion. i changed the interval...

$$\:\:{thank}\:{you}\:{for}\:{your}\:{attantion}. \\ $$$${i}\:{changed}\:{the}\:{interval}... \\ $$

Answered by Dwaipayan Shikari last updated on 13/Jan/21

(1/2)∫_0 ^(π/4) (dx/((cos((π/4)−x)+1)^2 ))=(1/2)∫_0 ^(π/4) (1/((1+cosx)^2 ))dx tan(x/2)=t =∫_0 ^((√2)−1) (1/((1+((1−t^2 )/(1+t^2 )))^2 )).(1/(1+t^2 ))dt =∫_0 ^((√2)−1) ((1+t^2 )/4)dt=(((√2)−1)/4)+((((√2)−1)^3 )/(12))=((2(√2))/3)−(5/6)

$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{dx}}{\left({cos}\left(\frac{\pi}{\mathrm{4}}−{x}\right)+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}}{\left(\mathrm{1}+{cosx}\right)^{\mathrm{2}} }{dx}\:\:\:\:\:\:\:{tan}\frac{{x}}{\mathrm{2}}={t} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} }.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{4}}{dt}=\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{4}}+\frac{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{12}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}−\frac{\mathrm{5}}{\mathrm{6}} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 13/Jan/21

very nice as always..thx...

$${very}\:{nice}\:{as}\:{always}..{thx}... \\ $$

Answered by MJS_new last updated on 13/Jan/21

∫(dx/((sin x +cos x +(√2))^2 ))= [t=x+((5π)/4) →dx=dt] =(1/2)∫(dt/((1−sin t)^2 ))= [u=tan (t/2) → dt=((2du)/(u^2 +1))] =∫((u^2 +1)/((u−1)^4 ))du=−((3u^2 −3u+2)/(3(u−1)^3 )) ⇒ answer is −(5/6)+((2(√2))/3)

$$\int\frac{{dx}}{\left(\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\:+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[{t}={x}+\frac{\mathrm{5}\pi}{\mathrm{4}}\:\rightarrow{dx}={dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{\left(\mathrm{1}−\mathrm{sin}\:{t}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[{u}=\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:\rightarrow\:{dt}=\frac{\mathrm{2}{du}}{{u}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\int\frac{{u}^{\mathrm{2}} +\mathrm{1}}{\left({u}−\mathrm{1}\right)^{\mathrm{4}} }{du}=−\frac{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{3}{u}+\mathrm{2}}{\mathrm{3}\left({u}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:−\frac{\mathrm{5}}{\mathrm{6}}+\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$

Commented by mnjuly1970 last updated on 13/Jan/21

thanks alot mr mjs.grateful..

$${thanks}\:{alot}\:{mr}\:{mjs}.{grateful}.. \\ $$

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