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 by maxmathsup 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=2x I=∫_0 ^π lnsint×(dt/2) I=∫_0 ^π lnsint×(dt/2) (1/2)∫_0 ^π lnsintdt [now ∫_0 ^(2a) f(x)dx=2∫_0 ^a f(x)dx when f(2a−x)=f(x)] so (1/2)∫_0 ^π lnsintdt [lnsin(π−t)=lnsint] I=(1/2)×2∫_0 ^(π/2) lnsintdt I=∫_0 ^(π/2) lnsin((π/2)−t)dt 2I=∫_0 ^(π/2) lnsint+lncost dt(/)p 2I=∫_0 ^(π/2) ln(((2sintcost)/2))dt =∫_0 ^(π/2) [lnsin2t−ln2]dt 2I=∫_0 ^(π/2) lnsin2tdt−ln2∫_0 ^(π/2) dt 2I=I−ln2×∣t∣_0 ^(π/2) I=−ln2×(π/2) pls check...](Q52965.png)
$${t}=\mathrm{2}{x} \\ $$$${I}=\int_{\mathrm{0}} ^{\pi} {lnsint}×\frac{{dt}}{\mathrm{2}} \\ $$$${I}=\int_{\mathrm{0}} ^{\pi} {lnsint}×\frac{{dt}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {lnsintdt} \\ $$$$\left[{now}\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:{when}\:{f}\left(\mathrm{2}{a}−{x}\right)={f}\left({x}\right)\right] \\ $$$${so}\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} {lnsintdt}\:\:\:\left[{lnsin}\left(\pi−{t}\right)={lnsint}\right] \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {lnsintdt} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {lnsin}\left(\frac{\pi}{\mathrm{2}}−{t}\right){dt} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {lnsint}+{lncost}\:{dt}\frac{}{}{p} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\frac{\mathrm{2}{sintcost}}{\mathrm{2}}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left[{lnsin}\mathrm{2}{t}−{ln}\mathrm{2}\right]{dt} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {lnsin}\mathrm{2}{tdt}−{ln}\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dt} \\ $$$$\mathrm{2}{I}={I}−{ln}\mathrm{2}×\mid{t}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$${I}=−{ln}\mathrm{2}×\frac{\pi}{\mathrm{2}}\:{pls}\:{check}... \\ $$
Commented by gunawan last updated on 16/Jan/19
$$\mathrm{Nice}\:\mathrm{Sir} \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19
$${most}\:{welcome}... \\ $$