Integration Questions

Question Number 151429 by peter frank last updated on 21/Aug/21

$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{xdx}}{\left(\mathrm{a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{b}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{2}} } \\$$

Answered by Olaf_Thorendsen last updated on 21/Aug/21

$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\pi} \frac{{x}}{\left({a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }\:{dx} \\$$$$\mathrm{Let}\:{u}\:=\:\pi−{x}\:: \\$$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\pi} \frac{\pi−{u}}{\left({a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {u}+{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {u}\right)^{\mathrm{2}} }\:{du} \\$$$$\Rightarrow\:\mathrm{2I}\:=\:\int_{\mathrm{0}} ^{\pi} \frac{\pi}{\left({a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {u}+{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {u}\right)^{\mathrm{2}} }\:{du} \\$$$$\Rightarrow\:\mathrm{I}\:=\:\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{{du}}{{a}^{\mathrm{4}} \mathrm{cos}^{\mathrm{4}} {u}\left(\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\mathrm{tan}^{\mathrm{2}} {u}\right)^{\mathrm{2}} } \\$$$$\:\mathrm{I}\:=\:\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{du}}{{a}^{\mathrm{4}} \mathrm{cos}^{\mathrm{4}} {u}\left(\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\mathrm{tan}^{\mathrm{2}} {u}\right)^{\mathrm{2}} } \\$$$$\mathrm{Let}\:{t}\:=\:\mathrm{tan}{u}\:: \\$$$$\mathrm{I}\:=\:\pi\int_{\mathrm{0}} ^{\infty} \frac{\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }}{{a}^{\mathrm{4}} \frac{\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\left(\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\$$$$\mathrm{I}\:=\:\frac{\pi}{{a}^{\mathrm{4}} }\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+{t}^{\mathrm{2}} }{\left(\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt} \\$$$$\mathrm{I}\:=\:\frac{\pi}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \frac{\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{t}^{\mathrm{2}} }{\left(\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt} \\$$$$\mathrm{I}\:=\:\frac{\pi}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{t}^{\mathrm{2}} }−\frac{\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}{\left(\mathrm{1}+\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\right)\:{dt} \\$$$$\mathrm{I}\:=\:\frac{\pi}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\left[\frac{{a}}{{b}}\mathrm{arctan}\left(\frac{{b}}{{a}}{t}\right)+\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}}.\frac{{t}}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} {t}^{\mathrm{2}} \right)}\right. \\$$$$\left.+\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ab}}\mathrm{arctan}\left(\frac{{b}}{{a}}{t}\right)\right]_{\mathrm{0}} ^{\infty} \\$$$$\mathrm{I}\:=\:\frac{\pi}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }\left(\frac{\pi}{\mathrm{2}}.\frac{{a}}{{b}}+\mathrm{0}+\frac{\pi}{\mathrm{2}}.\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ab}}\right) \\$$$$\mathrm{I}\:=\:\frac{\pi^{\mathrm{2}} \left(\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{\mathrm{4}{a}^{\mathrm{3}} {b}^{\mathrm{3}} } \\$$