Question Number 42232 by maxmathsup by imad last updated on 20/Aug/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{d}\theta}{\left(\mathrm{1}+{cos}\theta\right)^{\mathrm{3}} } \\ $$
Answered by MJS last updated on 21/Aug/18
$$\mathrm{simply}\:\mathrm{Weierstrass} \\ $$$${t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:\rightarrow\:{d}\theta=\frac{\mathrm{2}{td}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int\frac{{d}\theta}{\left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{4}}\int\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} {dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int{t}^{\mathrm{4}} {dt}+\frac{\mathrm{1}}{\mathrm{2}}\int{t}^{\mathrm{2}} {dt}+\frac{\mathrm{1}}{\mathrm{4}}\int{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{20}}{t}^{\mathrm{5}} +\frac{\mathrm{1}}{\mathrm{6}}{t}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{4}}{t}=\frac{\left(\mathrm{3}{t}^{\mathrm{4}} +\mathrm{10}{t}^{\mathrm{2}} +\mathrm{15}\right){t}}{\mathrm{60}}= \\ $$$$=\frac{\left(\mathrm{3tan}^{\mathrm{4}} \:\frac{{x}}{\mathrm{2}}\:+\mathrm{10tan}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\:+\mathrm{15}\right)\mathrm{tan}\:\frac{{x}}{\mathrm{2}}}{\mathrm{60}}+{C} \\ $$