Question Number 40141 by maxmathsup by imad last updated on 16/Jul/18
$${find}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{{dx}}{\mathrm{3}+{sinx}} \\ $$
Commented by maxmathsup by imad last updated on 19/Jul/18
![let I = ∫_0 ^(π/2) (dx/(3+sinx)) changement tan((x/2))=t give I = ∫_0 ^1 (1/(3+((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) = 2 ∫_0 ^1 (dt/(3+3t^2 +2t)) =2 ∫_0 ^1 (dt/(3t^2 +2t +3)) but 3t^2 +2t +3 =3(t^2 +(2/3)t +1) =3( t^2 +(2/3)t +(1/9) +1−(1/9)) =3{ (t+(1/3))^2 +(8/9)} ⇒ I =(2/3) ∫_0 ^1 (dt/((t+(1/3))^2 +(8/9))) =_(t+(1/3)=((2(√2))/3)u) (2/3) ∫_(1/(2(√2))) ^(√2) (1/((8/9)(1+u^2 ))) ((2(√2))/3)du =((4(√2))/9) (9/8) ∫_(1/(2(√2))) ^(√2) (du/(1+u^2 )) =((√2)/2) [arctanu]_(1/( (√2))) ^(√2) ⇒ I = ((√2)/2){ arctan((√2))−arctan((1/( (√2))))}.](https://www.tinkutara.com/question/Q40343.png)
$${let}\:\:{I}\:\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dx}}{\mathrm{3}+{sinx}}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\mathrm{3}+\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{2}{t}}\: \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\mathrm{3}{t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{3}}\:\:\:{but}\:\:\mathrm{3}{t}^{\mathrm{2}} +\mathrm{2}{t}\:+\mathrm{3}\:\:=\mathrm{3}\left({t}^{\mathrm{2}} \:+\frac{\mathrm{2}}{\mathrm{3}}{t}\:+\mathrm{1}\right) \\ $$$$=\mathrm{3}\left(\:{t}^{\mathrm{2}} \:+\frac{\mathrm{2}}{\mathrm{3}}{t}\:+\frac{\mathrm{1}}{\mathrm{9}}\:+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}\right)\:=\mathrm{3}\left\{\:\left({t}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \:+\frac{\mathrm{8}}{\mathrm{9}}\right\}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{2}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left({t}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \:+\frac{\mathrm{8}}{\mathrm{9}}}\:\:=_{{t}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}{u}} \:\:\:\frac{\mathrm{2}}{\mathrm{3}}\:\int_{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{\mathrm{1}}{\frac{\mathrm{8}}{\mathrm{9}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}{du} \\ $$$$=\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{9}}\:\frac{\mathrm{9}}{\mathrm{8}}\:\int_{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\left[{arctanu}\right]_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\sqrt{\mathrm{2}}} \:\:\Rightarrow \\ $$$${I}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left\{\:{arctan}\left(\sqrt{\mathrm{2}}\right)−{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\right\}. \\ $$