Question Number 83064 by M±th+et£s last updated on 27/Feb/20
$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sin}\left({nx}\right)}{{sin}\left({x}\right)}{dx}=\frac{\pi}{\mathrm{2}} \\ $$$${n}\:{is}\:{posative}\:{odd}\:{number} \\ $$
Answered by mind is power last updated on 27/Feb/20
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left(\mathrm{2}{kx}+{x}\right)}{{sin}\left({x}\right)}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left\{\frac{{sin}\left(\mathrm{2}{kx}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}+{cos}\left(\mathrm{2}{kx}\right)\right\}{dx} \\ $$$${k}=\mathrm{0}\:\:{evidente} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}\:−\mathrm{1}} {\sum}}{e}^{{ikx}} =\frac{\mathrm{1}−\left({e}^{{inx}} \right)}{\mathrm{1}−{e}^{{ix}} }=\frac{{e}^{{i}\left(\frac{{n}−\mathrm{1}}{\mathrm{2}}\right){x}} {sin}\left(\frac{{n}}{\mathrm{2}}{x}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{e}^{{i}\mathrm{2}{kx}} =\frac{{e}^{{i}\left({n}−\mathrm{1}\right){x}} {sin}\left({nx}\right)}{{sin}\left({x}\right)} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{e}^{{i}\left(\mathrm{2}{k}−{n}+\mathrm{1}\right){x}} =\frac{{sin}\left({nx}\right)}{{sin}\left({x}\right)} \\ $$$${n}=\mathrm{2}{m}+\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{m}} {\sum}}{e}^{{i}\left(\mathrm{2}{k}−\mathrm{2}{m}\right){x}} {dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left(\left(\mathrm{2}{m}+\mathrm{1}\right){x}\right)}{{sin}\left({x}\right)}{dx} \\ $$$${m}>\mathrm{0}\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \underset{{k}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}{e}^{{i}\left(\mathrm{2}{k}−\mathrm{2}{m}\right){x}} {dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{{i}.\left(\mathrm{2}{m}−\mathrm{2}{m}\right){x}} {dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \underset{{k}={m}+\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}{e}^{{i}\left(\mathrm{2}{k}−\mathrm{2}{m}\right){x}} {dx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{{i}\left(\mathrm{2}{k}−\mathrm{2}{m}\right){x}} {dx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\left[\frac{{e}^{{i}\left(\mathrm{2}{k}−\mathrm{2}{m}\right){x}} }{{i}\left(\mathrm{2}{k}−\mathrm{2}{m}\right)}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}+\underset{{m}+\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}\left[\frac{{e}^{{i}\left(\mathrm{2}{k}−\mathrm{2}{m}\right){x}} }{\mathrm{2}{k}−\mathrm{2}{m}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−{m}} −\mathrm{1}}{{i}\mathrm{2}\left({k}−{m}\right)}\:\:\:+\frac{\pi}{\mathrm{2}}+\underset{{m}+\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}\left[\frac{\left(−\mathrm{1}\right)^{{k}−{m}} −\mathrm{1}}{{i}\mathrm{2}\left({k}−{m}\right)}\right] \\ $$$$\underset{{m}+\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}\left[\frac{\left(−\mathrm{1}\right)^{{k}−{m}} −\mathrm{1}}{\mathrm{2}{i}\left({k}−{m}\right)}\right]=\underset{\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\left[\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{m}−{k}−{m}} −\mathrm{1}}{\mathrm{2}{i}\left(\mathrm{2}{m}−{k}−{m}\right)}\right] \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\left[\frac{\left(−\mathrm{1}\right)^{{k}−{m}} −\mathrm{1}}{\mathrm{2}{i}\left({m}−{k}\right)}\right]=−\underset{{k}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\left[\frac{\left(−\mathrm{1}\right)^{{k}−{m}} −\mathrm{1}}{\mathrm{2}{i}\left({k}−{m}\right)}\right] \\ $$$$\Rightarrow\underset{{m}+\mathrm{1}} {\overset{\mathrm{2}{m}} {\sum}}\left[\frac{\left(−\mathrm{1}\right)^{{k}−{m}} −\mathrm{1}}{\mathrm{2}{i}\left({k}−{m}\right)}\right]+\underset{{k}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\left[\frac{\left(−\mathrm{1}\right)^{{k}−{m}} −\mathrm{1}}{\mathrm{2}{i}\left({k}−{m}\right)}\right]=\mathrm{0} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\sum}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{{i}\mathrm{2}\left({k}−{m}\right){x}} {dx}=\frac{\pi}{\mathrm{2}}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\left({nx}\right)}{{sin}\left({x}\right)}{dx} \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 27/Feb/20
$${god}\:{bless}\:{you} \\ $$
Commented by mind is power last updated on 27/Feb/20
$${withe}\:{pleasur}\:{god}\:{bless}\:\:{evrey}\:{one}\: \\ $$$$ \\ $$
Answered by TANMAY PANACEA last updated on 28/Feb/20
$${I}_{{n}} −{I}_{{n}−\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sinnx}−{sin}\left({n}−\mathrm{2}\right){x}}{{sinx}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}{cos}\left({n}−\mathrm{1}\right){xsinx}}{{sinx}}{dx} \\ $$$${I}_{{n}} −{I}_{{n}−\mathrm{2}} =\mathrm{2}\mid\frac{{sin}\left({n}−\mathrm{1}\right){x}}{{n}−\mathrm{1}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\mathrm{2}}{{n}−\mathrm{1}}{sin}\left({n}−\mathrm{1}\right)\frac{\pi}{\mathrm{2}}=\mathrm{0} \\ $$$$\bigstar\:{when}\:{n}={odd}\:\:\:{then}\:{n}−\mathrm{1}={even} \\ $$$${so}\:{sin}\left({n}−\mathrm{1}\right)\frac{\pi}{\mathrm{2}}={sin}\left(\mathrm{2}{k}×\frac{\pi}{\mathrm{2}}\right)=\mathrm{0}\bigstar \\ $$$${I}_{{n}} ={I}_{{n}−\mathrm{2}} \: \\ $$$${I}_{\mathrm{3}} ={I}_{\mathrm{1}} \\ $$$${I}_{\mathrm{5}} ={I}_{\mathrm{3}} \\ $$$${so}\:{I}_{\mathrm{1}} ={I}_{\mathrm{3}} ={I}_{\mathrm{5}} ={I}_{\mathrm{7}} …. \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sinx}}{{sinx}}{dx}=\mid{x}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi}{\mathrm{2}} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{I}}_{{n}} =\frac{\pi}{\mathrm{2}} \\ $$
Commented by M±th+et£s last updated on 28/Feb/20
$${god}\:{bless}\:{you}\:{sir} \\ $$
Commented by TANMAY PANACEA last updated on 28/Feb/20
$${blessing}\:{shower}\:{to}\:{all} \\ $$