find-0-pi-dx-2-cosx-sinx-2-

Question Number 140976 by Mathspace last updated on 14/May/21

$${find}\:\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\left(\mathrm{2}−{cosx}−{sinx}\right)^{\mathrm{2}} } \\ $$$$ \\ $$

Answered by Ar Brandon last updated on 14/May/21

I=∫_0 ^π (dx/((2−cosx−sinx)^2 )) , t=tan(x/2) =∫_0 ^∞ ((2dt)/((2−((1−t^2 )/(1+t^2 ))−((2t)/(1+t^2 )))^2 ))∙(1/(1+t^2 ))=2∫_0 ^∞ ((t^2 +1)/((3t^2 −2t+1)^2 ))dt =(2/3)∫_0 ^∞ ((3t^2 −2t+1)/((3t^2 −2t+1)^2 ))dt+2∫_0 ^∞ ((2t/3+2/3)/((3t^2 −2t+1)^2 ))dt =(2/3)∫_0 ^∞ (dt/(3t^2 −2t+1))+(4/3)∫_0 ^∞ ((t+1)/((3t^2 −2t+1)^2 ))dt =(2/3)∫_0 ^∞ (dt/(3t^2 −2t+1))+[(4/3)∙((t−1/2)/(3t^2 −2t+1))+(4/3)∫(dt/(3t^2 −2t+1))]_0 ^∞ =2∫_0 ^∞ (dt/(3t^2 −2t+1))+(2/3)∙[((2t−1)/(3t^2 −2t+1))]_0 ^∞ =(2/3)∫_0 ^∞ (dt/(t^2 −(2/3)t+(1/3)))+(2/3)=(2/3)∫_0 ^∞ (dt/((t−(1/3))^2 +(2/9)))+(2/3) =(2/3)∙[(3/( (√2)))arctan(((3t−1)/( (√2))))]_0 ^∞ +(2/3)=(√2)((π/2)+tan^(−1) (((√2)/2)))+(2/3) ≈3.758527887

$$\mathcal{I}=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{dx}}{\left(\mathrm{2}−\mathrm{cosx}−\mathrm{sinx}\right)^{\mathrm{2}} }\:,\:\mathrm{t}=\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2dt}}{\left(\mathrm{2}−\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }−\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{3t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{3t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}}{\left(\mathrm{3t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt}+\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2t}/\mathrm{3}+\mathrm{2}/\mathrm{3}}{\left(\mathrm{3t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dt}}{\mathrm{3t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}}+\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}+\mathrm{1}}{\left(\mathrm{3t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dt}}{\mathrm{3t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}}+\left[\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{t}−\mathrm{1}/\mathrm{2}}{\mathrm{3t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}}+\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\mathrm{dt}}{\mathrm{3t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dt}}{\mathrm{3t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}}+\frac{\mathrm{2}}{\mathrm{3}}\centerdot\left[\frac{\mathrm{2t}−\mathrm{1}}{\mathrm{3t}^{\mathrm{2}} −\mathrm{2t}+\mathrm{1}}\right]_{\mathrm{0}} ^{\infty} \\ $$$$\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}\mathrm{t}+\frac{\mathrm{1}}{\mathrm{3}}}+\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dt}}{\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{9}}}+\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\centerdot\left[\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\mathrm{arctan}\left(\frac{\mathrm{3t}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{2}}{\mathrm{3}}=\sqrt{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\right)+\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\:\:\:\approx\mathrm{3}.\mathrm{758527887} \\ $$

Commented by Ar Brandon last updated on 14/May/21