Integration Questions

Question Number 153555 by mnjuly1970 last updated on 08/Sep/21

$$\\$$$$\:\:\:\:\:\Omega\::=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {cos}\left(\mathrm{2}{x}\right).{ln}\left({sin}\left({x}\right)\right){dx}\overset{?} {=}\:−\frac{\pi}{\mathrm{4}} \\$$$$\:\:\:\:\:\:\:\:\:\:{solution}\:\left(\mathrm{1}\:\right) \\$$$$\:\:\:\:\:\Omega\::=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left(\:\mathrm{2}{cos}^{\:\mathrm{2}} \left({x}\right)−\mathrm{1}\right){ln}\left({sin}\left({x}\right)\right){dx} \\$$$$\:\:\:\:\:\::=\:\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {cos}^{\:\mathrm{2}} \left({x}\right).{ln}\left({sin}\left({x}\right)\right){dx}−\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({x}\right)\right){dx} \\$$$$\:\:\:\:\:{we}\:{know}\:{that}\::\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({x}\right)\right){dx}\underset{{earlier}} {\overset{{derived}} {=}}\:\frac{−\pi}{\mathrm{2}}\:{ln}\left(\mathrm{2}\right) \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {cos}^{\:\mathrm{2}} \left({x}\right).{ln}\left({sin}\left({x}\right)\right){dx}\underset{{posts}} {\overset{{previous}} {=}}\:−\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{8}} \\$$$$\:\:\:\:\:\:\therefore\:\:\:\Omega\::=\:−\frac{\pi}{\mathrm{2}}\:{ln}\left(\mathrm{2}\right)\:−\frac{\pi}{\mathrm{4}}\:+\frac{\pi}{\mathrm{2}}\:{ln}\left(\mathrm{2}\right) \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacktriangleleft\:\:\:\:\Omega\:=−\:\frac{\pi}{\mathrm{4}}\:\:\blacktriangleright\:\:\:\:\:\:{m}.{n} \\$$

Answered by Ar Brandon last updated on 08/Sep/21

$$\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\left(\mathrm{2}{x}\right)\mathrm{ln}\left(\mathrm{sin}{x}\right){dx} \\$$$$\begin{cases}{{u}\left({x}\right)=\mathrm{ln}\left(\mathrm{sin}{x}\right)}\\{{v}'\left({x}\right)=\mathrm{cos}\left(\mathrm{2}{x}\right)}\end{cases}\Rightarrow\begin{cases}{{u}'\left({x}\right)=\frac{\mathrm{cos}{x}}{\mathrm{sin}{x}}}\\{{v}\left({x}\right)=\frac{\mathrm{sin2}{x}}{\mathrm{2}}}\end{cases} \\$$$$\Omega=\left[\frac{\mathrm{ln}\left(\mathrm{sin}{x}\right)\mathrm{sin2}{x}}{\mathrm{2}}−\int\mathrm{cos}^{\mathrm{2}} {xdx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$$$\:\:\:\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} {xdx}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+\mathrm{cos2}{x}\right){dx} \\$$$$\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\left[{x}+\frac{\mathrm{sin2}{x}}{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =−\frac{\pi}{\mathrm{4}} \\$$

Commented by mnjuly1970 last updated on 08/Sep/21

$$\:{very}\:{nice}\:{sir}\:{brando}\:{as}\:{always}.. \\$$

Commented by Ar Brandon last updated on 08/Sep/21

Thanks