Question Number 53418 by Abdo msup. last updated on 21/Jan/19
$${find}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{x}}{\mathrm{2}+{cosx}\:{sinx}}{dx} \\ $$
Commented by maxmathsup by imad last updated on 22/Jan/19
![let I =∫_0 ^π (x/(2 +cosx sinx))dx ⇒I =∫_0 ^π (x/(2+(1/2)sin(2x)))dx =∫_0 ^π ((2xdx)/(4+sin(2x))) =_(2x=t) ∫_0 ^(2π) ((t )/(4+sin(t))) (dt/2) =(1/2) ∫_0 ^(2π) (t/(4+sint))dt ⇒ 2I = ∫_0 ^π (t/(4+sin(t)))dt +∫_π ^(2π) (t/(4+sin(t)))dt =H+K K =_(t=π +u) ∫_0 ^π ((π+u)/(4−sinu)) du let find H H =_(tan((t/2))=u) ∫_0 ^∞ ((2arctanu)/(4+((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) =∫_0 ^∞ ((4 arctan(u))/(4+4u^2 +2u)) du =2 ∫_0 ^∞ ((arctan(u))/(2u^2 +u +2)) du =∫_0 ^∞ ((arctan(u))/(u^2 +(u/(2 ))+1))du =∫_0 ^∞ ((arctan(u))/(u^2 +2(1/4)u +(1/(16))+1−(1/(16))))du =∫_0 ^∞ ((arctan(u))/((u+(1/4))^2 +((15)/(16))))du =_(u+(1/4)=((√(15))/4)α) ((16)/(15))∫_(1/(√(15))) ^∞ ((artan(((α(√(15))−1)/4)))/(1+α^2 )) ((√(15))/4) dα =(4/(√(15)))∫_(1/(√(15))) ^∞ ((arctan(((α(√(15))−1)/4)))/(1+α^2 )) dα this integral is at at form ∫_λ ^∞ ((arctan(ax+b))/(x^2 +1))dx wich want a special studies... ( you can consider f(a)=∫_λ ^∞ ((arctan(ax+b))/(x^2 +1))dx and calculate first f^′ (a)...) we have K =π ∫_0 ^π (du/(4−sinu)) +∫_0 ^π ((udu)/(4−sinu)) ∫_0 ^π ((udu)/(4−sinu)) drive us to ∫_0 ^∞ ((arctan(ax+b))/(x^2 +1)) dx.... let determine ∫_0 ^π (du/(4−sinu)) =_(tan((u/2))=t) ∫_0 ^∞ (1/(4−((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^∞ ((2dt)/(4+4t^2 −2t)) =∫_0 ^∞ (dt/(2t^2 −t +2)) dt =(1/2)∫_0 ^∞ (dt/(t^2 −(t/2)+1)) =(1/2)∫_0 ^∞ (dt/(t^2 −2(1/4)t +(1/(16))+1−(1/(16)))) =(1/2)∫_0 ^∞ (dt/((t−(1/4))^2 +((15)/(16)))) =_(t−(1/4)=((√(15))/4)u) (1/2) ((16)/(15)) ∫_(−(1/(√(15)))) ^(+∞) (1/(1+u^2 )) ((√(15))/4) du =(2/(√(15)))[arctan(u)]_(−(1/(√(15)))) ^(+∞) =(2/(√(15))) {(π/2) +arctan((1/(√(15))))}](Q53456.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{x}}{\mathrm{2}\:+{cosx}\:{sinx}}{dx}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{x}}{\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{2}{xdx}}{\mathrm{4}+{sin}\left(\mathrm{2}{x}\right)}\:=_{\mathrm{2}{x}={t}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{t}\:}{\mathrm{4}+{sin}\left({t}\right)}\:\frac{{dt}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{t}}{\mathrm{4}+{sint}}{dt}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{t}}{\mathrm{4}+{sin}\left({t}\right)}{dt}\:+\int_{\pi} ^{\mathrm{2}\pi} \:\:\frac{{t}}{\mathrm{4}+{sin}\left({t}\right)}{dt}\:={H}+{K} \\ $$$${K}\:=_{{t}=\pi\:+{u}} \:\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{\pi+{u}}{\mathrm{4}−{sinu}}\:{du}\:\:\:{let}\:{find}\:{H}\: \\ $$$${H}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{arctanu}}{\mathrm{4}+\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{4}\:{arctan}\left({u}\right)}{\mathrm{4}+\mathrm{4}{u}^{\mathrm{2}} \:+\mathrm{2}{u}}\:{du} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({u}\right)}{\mathrm{2}{u}^{\mathrm{2}} \:+{u}\:+\mathrm{2}}\:{du}\:\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({u}\right)}{{u}^{\mathrm{2}} \:+\frac{{u}}{\mathrm{2}\:}+\mathrm{1}}{du}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({u}\right)}{{u}^{\mathrm{2}} \:+\mathrm{2}\frac{\mathrm{1}}{\mathrm{4}}{u}\:+\frac{\mathrm{1}}{\mathrm{16}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}}{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({u}\right)}{\left({u}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} \:+\frac{\mathrm{15}}{\mathrm{16}}}{du}\:=_{{u}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}\alpha} \:\:\:\:\frac{\mathrm{16}}{\mathrm{15}}\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{15}}}} ^{\infty} \:\:\:\frac{{artan}\left(\frac{\alpha\sqrt{\mathrm{15}}−\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{1}+\alpha^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}\:{d}\alpha \\ $$$$=\frac{\mathrm{4}}{\sqrt{\mathrm{15}}}\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{15}}}} ^{\infty} \:\:\:\:\frac{{arctan}\left(\frac{\alpha\sqrt{\mathrm{15}}−\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{1}+\alpha^{\mathrm{2}} }\:{d}\alpha\:\:{this}\:{integral}\:{is}\:{at}\:{at}\:{form} \\ $$$$\int_{\lambda} ^{\infty} \:\:\frac{{arctan}\left({ax}+{b}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:{wich}\:{want}\:{a}\:{special}\:{studies}... \\ $$$$\left(\:{you}\:{can}\:{consider}\:{f}\left({a}\right)=\int_{\lambda} ^{\infty} \:\:\frac{{arctan}\left({ax}+{b}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:{and}\:{calculate}\:{first}\:{f}^{'} \left({a}\right)...\right) \\ $$$${we}\:{have}\:{K}\:=\pi\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{du}}{\mathrm{4}−{sinu}}\:+\int_{\mathrm{0}} ^{\pi} \:\frac{{udu}}{\mathrm{4}−{sinu}} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\frac{{udu}}{\mathrm{4}−{sinu}}\:{drive}\:{us}\:{to}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({ax}+{b}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:{dx}....\:{let}\:{determine} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\frac{{du}}{\mathrm{4}−{sinu}}\:=_{{tan}\left(\frac{{u}}{\mathrm{2}}\right)={t}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{4}−\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\mathrm{4}+\mathrm{4}{t}^{\mathrm{2}} −\mathrm{2}{t}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dt}}{\mathrm{2}{t}^{\mathrm{2}} −{t}\:+\mathrm{2}}\:{dt}\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} −\frac{{t}}{\mathrm{2}}+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{2}\frac{\mathrm{1}}{\mathrm{4}}{t}\:+\frac{\mathrm{1}}{\mathrm{16}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} \:+\frac{\mathrm{15}}{\mathrm{16}}}\:=_{{t}−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}{u}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{16}}{\mathrm{15}}\:\int_{−\frac{\mathrm{1}}{\sqrt{\mathrm{15}}}} ^{+\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{15}}}{\mathrm{4}}\:{du} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{15}}}\left[{arctan}\left({u}\right)\right]_{−\frac{\mathrm{1}}{\sqrt{\mathrm{15}}}} ^{+\infty} \:=\frac{\mathrm{2}}{\sqrt{\mathrm{15}}}\:\left\{\frac{\pi}{\mathrm{2}}\:+{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{15}}}\right)\right\} \\ $$$$ \\ $$