Question Number 44422 by peter frank last updated on 28/Sep/18
$${use}\:{substitution}\:{x}=\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{3}{sin}^{\mathrm{2}} \theta \\ $$$${show}\:{that}\int_{\mathrm{1}} ^{\mathrm{3}} \frac{{dx}}{\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{3}−{x}\right)}}=\pi \\ $$
Commented by maxmathsup by imad last updated on 28/Sep/18
$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{−\mathrm{2}{cos}\theta\:{sin}\theta\:+\mathrm{6}{sin}\theta\:{cos}\theta}{\sqrt{\left({cos}^{\mathrm{2}} \theta−\mathrm{1}+\mathrm{3}{sin}^{\mathrm{2}} \theta\right)\left(\mathrm{3}−\mathrm{3}{sin}^{\mathrm{2}} \theta−{cos}^{\mathrm{2}} \theta\right)}}{d}\theta \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{4}{sin}\theta\:.{cos}\theta}{\sqrt{\mathrm{2}{sin}^{\mathrm{2}} \theta.\mathrm{2}{cos}^{\mathrm{2}} \theta}}{d}\theta\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{d}\theta\:=\pi\:. \\ $$
Commented by peter frank last updated on 29/Sep/18
$${thank}\:{you} \\ $$
Commented by maxmathsup by imad last updated on 29/Sep/18
$${you}\:{are}\:{welcome}\:{sir}. \\ $$