Question Number 85677 by john santu last updated on 24/Mar/20
$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:\left(\frac{\mathrm{21}{x}}{\mathrm{2}}\right)}{\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)}\:{dx}\: \\ $$
Commented by jagoll last updated on 24/Mar/20
$$\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{way}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\mathrm{it} \\ $$
Commented by john santu last updated on 24/Mar/20
$${using}\:{Reduction}\:{formula} \\ $$
Commented by M±th+et£s last updated on 25/Mar/20
![∫_0 ^π ((sin(((21x)/2)))/(sin((x/2))))dx let y=(x/2) ∫_0 ^(π/2) (((sin21y))/(sin(y))) 2dy 2∫_0 ^(π/2) ((sin20y cos(y) + cos20y siny)/(siny))dy 2∫_0 ^(π/2) sin20y coty dy +∫_0 ^(π/2) cos20y dy ∫_0 ^(π/2) cosy dy=0 2∫_0 ^(π/2) sin20y coty dy=2∫_0 ^(π/2) (1+2cos2y+2cos4y+...+cos20y)dy [y+sin2y+(1/2)sin4y+(1/3)sin6y+...]_0 ^(π/2) π+0+.... so ∫_0 ^π ((sin(((21x)/2)))/(sin((x/2))))dx=π](https://www.tinkutara.com/question/Q85844.png)
$$\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\frac{\mathrm{21}{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{dx} \\ $$$${let}\:{y}=\frac{{x}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left({sin}\mathrm{21}{y}\right)}{{sin}\left({y}\right)}\:\mathrm{2}{dy} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}\mathrm{20}{y}\:{cos}\left({y}\right)\:+\:{cos}\mathrm{20}{y}\:{siny}}{{siny}}{dy} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}\mathrm{20}{y}\:\:{coty}\:{dy}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\mathrm{20}{y}\:{dy} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cosy}\:{dy}=\mathrm{0} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}\mathrm{20}{y}\:{coty}\:{dy}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+\mathrm{2}{cos}\mathrm{2}{y}+\mathrm{2}{cos}\mathrm{4}{y}+…+{cos}\mathrm{20}{y}\right){dy} \\ $$$$ \\ $$$$\left[{y}+{sin}\mathrm{2}{y}+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{4}{y}+\frac{\mathrm{1}}{\mathrm{3}}{sin}\mathrm{6}{y}+…\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\pi+\mathrm{0}+…. \\ $$$${so}\:\:\:\:\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\frac{\mathrm{21}{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{dx}=\pi \\ $$
Answered by mind is power last updated on 24/Mar/20
![Σ_(k=0) ^(20) e^(ikx) =((e^(i10x) sin(((21x)/2)))/(sin((x/2)))) ⇒∫_0 ^π ((sin(((21x)/2)))/(sin((x/2))))dx=Σ_(k.0) ^(20) ∫_0 ^π e^(i(k−10)x) dx easy now =π+Σ_(k=0) ^9 ∫_0 ^π (e^(i(k−10)x) )dx+Σ_(k=0) ^9 ∫_0 ^π (e^(i(1+k)x) )dx =π+Σ_(k=0) ^9 {(1/(i(k−10)))[e^(i(k−10)π) −1]+((e^(i(1+k)π) −1)/(i(1+k)))} =π+Σ_(k=0) ^9 (((−1)^k −1)/(i(k−10)))+Σ_(k=0) ^9 (((−1)^k −1)/(i(1+k))) =π−Σ_(k=0) ^9 (((−1)^k −1)/(i(1+k)))+Σ_(k=0) ^9 (((−1)^k −1)/(i(1+k)))=π](https://www.tinkutara.com/question/Q85765.png)
$$\underset{{k}=\mathrm{0}} {\overset{\mathrm{20}} {\sum}}{e}^{{ikx}} =\frac{{e}^{{i}\mathrm{10}{x}} \:{sin}\left(\frac{\mathrm{21}{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\frac{\mathrm{21}{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{dx}=\underset{{k}.\mathrm{0}} {\overset{\mathrm{20}} {\sum}}\int_{\mathrm{0}} ^{\pi} {e}^{{i}\left({k}−\mathrm{10}\right){x}} {dx}\:{easy}\:{now} \\ $$$$=\pi+\underset{{k}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\int_{\mathrm{0}} ^{\pi} \left({e}^{{i}\left({k}−\mathrm{10}\right){x}} \right){dx}+\underset{{k}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\int_{\mathrm{0}} ^{\pi} \left({e}^{{i}\left(\mathrm{1}+{k}\right){x}} \right){dx} \\ $$$$=\pi+\underset{{k}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\left\{\frac{\mathrm{1}}{{i}\left({k}−\mathrm{10}\right)}\left[{e}^{{i}\left({k}−\mathrm{10}\right)\pi} −\mathrm{1}\right]+\frac{{e}^{{i}\left(\mathrm{1}+{k}\right)\pi} −\mathrm{1}}{{i}\left(\mathrm{1}+{k}\right)}\right\} \\ $$$$=\pi+\underset{{k}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} −\mathrm{1}}{{i}\left({k}−\mathrm{10}\right)}+\underset{{k}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} −\mathrm{1}}{{i}\left(\mathrm{1}+{k}\right)} \\ $$$$=\pi−\underset{{k}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} −\mathrm{1}}{{i}\left(\mathrm{1}+{k}\right)}+\underset{{k}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} −\mathrm{1}}{{i}\left(\mathrm{1}+{k}\right)}=\pi \\ $$
Commented by jagoll last updated on 25/Mar/20
$$\mathrm{sir}\:\mathrm{what}\:\mathrm{it}\:\mathrm{formula}? \\ $$
Commented by mind is power last updated on 26/Mar/20
$${Formula}\:? \\ $$
Answered by john santu last updated on 25/Mar/20
![I = ∫_0 ^π ((sin (((21x)/2)))/(sin ((x/2)))) dx, [ t = (x/2)] I = 2∫_0 ^(π/2) ((sin 21t)/(sin t)) dt = consider G_n = ∫_0 ^(π/2) ((sin nt)/(sin t)) dt then G_(n+2) −G_n = ∫_0 ^(π/2) ((sin ((n+2)t)−sin (nt))/(sin t)) dt = ∫ _0 ^(π/2) ((sin t cos ((n+1)t))/(sin t)) dt = ((sin ((((n+1)π)/2)))/(n+1)) G_(n+2) = G_n + ((sin ((((n+1)π)/2)))/(n+1)) , n≥ 0 when n is odd n = 2k−1 G_(2k+1) = G_(2k−1) , k≥ 1 G_(21) = G_1 = ∫^(π/2) _0 ((sin t)/(sin t)) dt = (π/2) so that I = ∫_0 ^π ((sin (((21x)/2)))/(sin ((x/2)))) dx = 2G_(21) =2×(π/2) = π](https://www.tinkutara.com/question/Q85862.png)
$${I}\:=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:\left(\frac{\mathrm{21}{x}}{\mathrm{2}}\right)}{\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)}\:{dx},\:\:\left[\:{t}\:=\:\frac{{x}}{\mathrm{2}}\right] \\ $$$${I}\:=\:\mathrm{2}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{sin}\:\mathrm{21}{t}}{\mathrm{sin}\:{t}}\:{dt}\:=\: \\ $$$${consider}\:{G}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{sin}\:{nt}}{\mathrm{sin}\:{t}}\:{dt}\: \\ $$$${then}\:{G}_{{n}+\mathrm{2}} −{G}_{{n}} \:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{sin}\:\left(\left({n}+\mathrm{2}\right){t}\right)−\mathrm{sin}\:\left({nt}\right)}{\mathrm{sin}\:{t}}\:{dt}\: \\ $$$$=\:\int\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\:}}\:\frac{\mathrm{sin}\:{t}\:\mathrm{cos}\:\left(\left({n}+\mathrm{1}\right){t}\right)}{\mathrm{sin}\:{t}}\:{dt}\: \\ $$$$=\:\frac{\mathrm{sin}\:\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)}{{n}+\mathrm{1}} \\ $$$${G}_{{n}+\mathrm{2}} \:=\:{G}_{{n}} +\:\frac{\mathrm{sin}\:\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)}{{n}+\mathrm{1}}\:,\:{n}\geqslant\:\mathrm{0} \\ $$$${when}\:{n}\:{is}\:{odd}\:{n}\:=\:\mathrm{2}{k}−\mathrm{1} \\ $$$${G}_{\mathrm{2}{k}+\mathrm{1}} \:=\:{G}_{\mathrm{2}{k}−\mathrm{1}} \:,\:{k}\geqslant\:\mathrm{1} \\ $$$${G}_{\mathrm{21}} \:=\:{G}_{\mathrm{1}} \:=\:\underset{\mathrm{0}} {\int}^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}\:{t}}{\mathrm{sin}\:{t}}\:{dt}\:=\:\frac{\pi}{\mathrm{2}} \\ $$$${so}\:{that}\: \\ $$$${I}\:=\:\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:\left(\frac{\mathrm{21}{x}}{\mathrm{2}}\right)}{\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)}\:{dx}\:=\:\mathrm{2}{G}_{\mathrm{21}} \:=\mathrm{2}×\frac{\pi}{\mathrm{2}}\: \\ $$$$=\:\pi \\ $$