# 0-pi-2-cos-2-t-sint-dt-

Question Number 142643 by ArielVyny last updated on 03/Jun/21
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{2}} {t}}{{sint}}{dt} \\$$
Answered by MJS_new last updated on 03/Jun/21
$$\int\frac{\mathrm{cos}^{\mathrm{2}} \:{t}}{\mathrm{sin}\:{t}}{dt}=\int\frac{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{t}}{\mathrm{sin}\:{t}}{dt}=\int\left(−\mathrm{sin}\:{t}\:+\mathrm{csc}\:{t}\right){dt}= \\$$$$=\mathrm{cos}\:{t}\:+\mathrm{ln}\:\mathrm{tan}\:\frac{{t}}{\mathrm{2}}\:+{C} \\$$$$\Rightarrow\:\mathrm{Integral}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{converge} \\$$
Commented by ArielVyny last updated on 03/Jun/21
$${in}\:\left[\mathrm{0}.\frac{\pi}{\mathrm{2}}\right]\:{integral}\:{does}\:{not}\:{converge}\:? \\$$
Commented by MJS_new last updated on 03/Jun/21
$$\mathrm{yes}. \\$$$$\left[\mathrm{cos}\:{x}\overset{\pi/\mathrm{2}} {\right]}_{\mathrm{0}} =\mathrm{0}−\mathrm{1}=−\mathrm{1} \\$$$$\left[\mathrm{ln}\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} =\mathrm{0}−\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\mathrm{ln}\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\:\:\:\:\mathrm{but}\:\mathrm{this}\:\mathrm{limit}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\$$