Question Number 168477 by infinityaction last updated on 11/Apr/22
Commented by mr W last updated on 11/Apr/22
$$\pi/\mathrm{3}>\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos}^{−\mathrm{1}} \left(\pi/\mathrm{3}\right)\:{is}\:{not}\:{defined}! \\ $$$${question}\:{is}\:{wrong}! \\ $$
Commented by infinityaction last updated on 11/Apr/22
$${no} \\ $$$${use}\:{of}\:{limit}\:{after}\:{integration} \\ $$
Commented by mr W last updated on 11/Apr/22
$${so}\:{you}\:{think}\:\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:{dx}\:{exists}? \\ $$
Commented by infinityaction last updated on 11/Apr/22
$${no} \\ $$$${ggot}\:{it}\:{sir} \\ $$
Commented by safojontoshtemirov last updated on 12/Apr/22
$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{3}}} {\int}}{sinx}\centerdot{cosxdx}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{3}}} {\int}}{sinxd}\left({sinx}\right)=\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} {\int}}{tdt} \\ $$$$\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}} {\mid}}=\frac{\mathrm{3}}{\mathrm{8}}. \\ $$
Commented by mr W last updated on 12/Apr/22
$$\mathrm{cos}^{−\mathrm{1}} {x}=\mathrm{arccos}\:{x}\neq\frac{\mathrm{1}}{\mathrm{cos}\:{x}} \\ $$