UNKNOWN Questions

Question Number 52946 by gunawan last updated on 15/Jan/19

$$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\:\mathrm{log}\:\mathrm{sin}\:\mathrm{2}{x}\:{dx}\:= \\$$

Commented bymaxmathsup by imad last updated on 15/Jan/19

$${let}\:{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sin}\left(\mathrm{2}{x}\right)\right){dx}\:\Rightarrow{A}=_{\mathrm{2}{x}={t}} \:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {log}\left({sint}\right){dt} \\$$ $$\Rightarrow\mathrm{2}{A}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sint}\right){dt}\:+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} {log}\left({sint}\right){dt}\:{but}\:{we}\:{have}\:{proved}\:{that} \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sint}\right){dt}\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:{also}\:\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} {log}\left({sint}\right){dt}\:=_{{t}\:=\frac{\pi}{\mathrm{2}}+{u}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({sin}\left(\frac{\pi}{\mathrm{2}}+{u}\right)\right){du} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {log}\left({cosu}\right){du}\:=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:\left({result}\:{proved}\right)\:\Rightarrow \\$$ $$\mathrm{2}{A}\:=−\pi{ln}\left(\mathrm{2}\right)\:\Rightarrow{A}=−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right). \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jan/19

$${t}=\mathrm{2}{x} \\$$ $${I}=\int_{\mathrm{0}} ^{\pi} {lnsint}×\frac{{dt}}{\mathrm{2}} \\$$ $${I}=\int_{\mathrm{0}} ^{\pi} {lnsint}×\frac{{dt}}{\mathrm{2}} \\$$ $$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {lnsintdt} \\$$ $$\left[{now}\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:{when}\:{f}\left(\mathrm{2}{a}−{x}\right)={f}\left({x}\right)\right] \\$$ $${so}\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {lnsintdt}\:\:\:\left[{lnsin}\left(\pi−{t}\right)={lnsint}\right] \\$$ $${I}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {lnsintdt} \\$$ $${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {lnsin}\left(\frac{\pi}{\mathrm{2}}−{t}\right){dt} \\$$ $$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {lnsint}+{lncost}\:{dt}\frac{}{}{p} \\$$ $$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{2}{sintcost}}{\mathrm{2}}\right){dt} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[{lnsin}\mathrm{2}{t}−{ln}\mathrm{2}\right]{dt} \\$$ $$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {lnsin}\mathrm{2}{tdt}−{ln}\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dt} \\$$ $$\mathrm{2}{I}={I}−{ln}\mathrm{2}×\mid{t}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$ $${I}=−{ln}\mathrm{2}×\frac{\pi}{\mathrm{2}}\:{pls}\:{check}... \\$$

Commented bygunawan last updated on 16/Jan/19

$$\mathrm{Nice}\:\mathrm{Sir} \\$$ $$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\$$

Commented bytanmay.chaudhury50@gmail.com last updated on 16/Jan/19

$${most}\:{welcome}... \\$$