UNKNOWN Questions

Question Number 53694 by gunawan last updated on 25/Jan/19

$$\:\underset{\:\mathrm{0}} {\overset{{r}\:\pi} {\int}}\:\:\mathrm{sin}^{\mathrm{2}{n}} {x}\:{dx}\:= \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

$$\int_{\mathrm{0}} ^{\pi} {sinxdx}=\mid−{cosx}\mid_{\mathrm{0}} ^{\pi} =−\left({cos}\pi−{cos}\mathrm{0}\right)=\mathrm{2} \\$$$${so}\:{the}\:{area}\:{of}\:{each}\:{loop}\:{of}\:{f}\left({x}\right)={sinx}\:{is}\:\mathrm{2} \\$$$${f}\left({x}\right)={sin}^{\mathrm{2}{n}} {x} \\$$$${f}\left({r}\pi−{x}\right)=\left[{sin}\left({r}\pi−{x}\right)\right]^{\mathrm{2}{n}} =\left[{eitther}\:+{sinx}\:{or}\:−{sinx}\right] \\$$$${but}\:\left[{sin}\left({r}\pi−{x}\right)\right]^{\mathrm{2}{n}} ={sin}^{\mathrm{2}{n}} {x} \\$$$${now}\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:{when}\:{f}\left(\mathrm{2}{a}−{x}\right)={f}\left({x}\right) \\$$$${sinx}\:={sin}\left(\mathrm{2}\pi+{x}\right)\:{peridocity}\:\mathrm{2}\pi \\$$$$\int_{\mathrm{0}} ^{{r}\pi} {sin}^{\mathrm{2}{n}} {xdx} \\$$$$=\int_{\mathrm{0}} ^{\frac{{r}}{\mathrm{2}}×\mathrm{2}\pi} {sin}^{\mathrm{2}{n}} {xdx} \\$$$$=\frac{{r}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} {sin}^{\mathrm{2}{n}} {xdx} \\$$$$=\frac{{r}}{\mathrm{2}}×\mathrm{2}\int_{\mathrm{0}} ^{\pi} {sin}^{\mathrm{2}{n}} {xdx}\left[\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:\:\:{whenf}\left(\mathrm{2}{a}−{x}\right)={f}\left({x}\right)\right] \\$$$$=\mathrm{2}{r}×\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}} {xdx} \\$$$${gamma}\:{function}... \\$$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} {xcos}^{\mathrm{2}{q}−\mathrm{1}} {xdx}=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)} \\$$$${r}×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}+\mathrm{1}−\mathrm{1}} {xcos}^{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dx} \\$$$$={r}×\frac{\lceil\left(\mathrm{2}{n}+\mathrm{1}\right)\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\mathrm{2}{n}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\$$$${i}\:{have}\:{tried}\:{to}\:{solve}..{others}\:{pls}\:{check}... \\$$