Question Number 32354 by abdo imad last updated on 23/Mar/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{sin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }\:. \\ $$
Commented by abdo imad last updated on 28/Mar/18
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dt}}{\left(\mathrm{1}+{sin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} } \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{dt}}{\left(\mathrm{1}+\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right)^{\mathrm{2}} }\:=\:\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\frac{{dt}}{\left(\mathrm{3}−{cos}\left(\mathrm{2}{t}\right)\right)^{\mathrm{2}} } \\ $$$$=_{\mathrm{2}{t}={x}} \:\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}−{cosx}\right)^{\mathrm{2}} }\:\frac{{dx}}{\mathrm{2}}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dx}}{\left(\mathrm{3}−{cosx}\right)^{\mathrm{2}} } \\ $$$${ch}.\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:={tgive} \\ $$$${I}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}\:−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\mathrm{4}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:\frac{\left(\mathrm{3}\:+\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$=\:\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\left(\mathrm{2}+\mathrm{4}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)}\:{dt}\:\:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} }{dt}\:=_{{t}\sqrt{\mathrm{2}}\:={u}} \:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\:\frac{{du}}{\:\sqrt{\mathrm{2}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{artan}\left(\sqrt{\mathrm{2}}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=_{{t}\sqrt{\mathrm{2}}={u}} \:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\frac{{du}}{\:\sqrt{\mathrm{2}}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{du}\:\:{but}\:\:\:\:\:\: \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{du}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:{u}.\left(\frac{−\mathrm{2}{u}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\right){du}\:{by}\:{parts} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\left(\:\left[\:\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\:{du}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:=\frac{−\sqrt{\mathrm{2}}}{\mathrm{6}}\:\:+\frac{\mathrm{1}}{\mathrm{4}}\:\left[{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \\ $$$$=\:\frac{−\sqrt{\mathrm{2}}}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{3}\right)\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\frac{−\sqrt{\mathrm{2}}}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{4}}{ln}\mathrm{3}\right) \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{12}}\:+\frac{{ln}\mathrm{3}}{\mathrm{8}\sqrt{\mathrm{2}}} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{arctan}\left(\sqrt{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{12}}\:−\frac{{ln}\mathrm{3}}{\mathrm{8}\sqrt{\mathrm{2}}}\:. \\ $$