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
![∫_0 ^π sinxdx=∣−cosx∣_0 ^π =−(cosπ−cos0)=2 so the area of each loop of f(x)=sinx is 2 f(x)=sin^(2n) x f(rπ−x)=[sin(rπ−x)]^(2n) =[eitther +sinx or −sinx] but [sin(rπ−x)]^(2n) =sin^(2n) x now ∫_0 ^(2a) f(x)dx=2∫_0 ^a f(x)dx when f(2a−x)=f(x) sinx =sin(2π+x) peridocity 2π ∫_0 ^(rπ) sin^(2n) xdx =∫_0 ^((r/2)×2π) sin^(2n) xdx =(r/2)∫_0 ^(2π) sin^(2n) xdx =(r/2)×2∫_0 ^π sin^(2n) xdx[∫_0 ^(2a) f(x)dx=2∫_0 ^a f(x)dx whenf(2a−x)=f(x)] =2r×∫_0 ^(π/2) sin^(2n) xdx gamma function... 2∫_0 ^(π/2) sin^(2p−1) xcos^(2q−1) xdx=((⌈(p)⌈(q))/(⌈(p+q))) r×2∫_0 ^(π/2) sin^(2n+1−1) xcos^(2×(1/2)−1) dx =r×((⌈(2n+1)⌈((1/2)))/(⌈(2n+1+(1/2)))) i have tried to solve..others pls check...](https://www.tinkutara.com/question/Q53718.png)
$$\int_{\mathrm{0}} ^{\pi} {sinxdx}=\mid−{cosx}\mid_{\mathrm{0}} ^{\pi} =−\left({cos}\pi−{cos}\mathrm{0}\right)=\mathrm{2} \\ $$$${so}\:{the}\:{area}\:{of}\:{each}\:{loop}\:{of}\:{f}\left({x}\right)={sinx}\:{is}\:\mathrm{2} \\ $$$${f}\left({x}\right)={sin}^{\mathrm{2}{n}} {x} \\ $$$${f}\left({r}\pi−{x}\right)=\left[{sin}\left({r}\pi−{x}\right)\right]^{\mathrm{2}{n}} =\left[{eitther}\:+{sinx}\:{or}\:−{sinx}\right] \\ $$$${but}\:\left[{sin}\left({r}\pi−{x}\right)\right]^{\mathrm{2}{n}} ={sin}^{\mathrm{2}{n}} {x} \\ $$$${now}\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:{when}\:{f}\left(\mathrm{2}{a}−{x}\right)={f}\left({x}\right) \\ $$$${sinx}\:={sin}\left(\mathrm{2}\pi+{x}\right)\:{peridocity}\:\mathrm{2}\pi \\ $$$$\int_{\mathrm{0}} ^{{r}\pi} {sin}^{\mathrm{2}{n}} {xdx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{{r}}{\mathrm{2}}×\mathrm{2}\pi} {sin}^{\mathrm{2}{n}} {xdx} \\ $$$$=\frac{{r}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} {sin}^{\mathrm{2}{n}} {xdx} \\ $$$$=\frac{{r}}{\mathrm{2}}×\mathrm{2}\int_{\mathrm{0}} ^{\pi} {sin}^{\mathrm{2}{n}} {xdx}\left[\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:\:\:{whenf}\left(\mathrm{2}{a}−{x}\right)={f}\left({x}\right)\right] \\ $$$$=\mathrm{2}{r}×\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}} {xdx} \\ $$$${gamma}\:{function}… \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} {xcos}^{\mathrm{2}{q}−\mathrm{1}} {xdx}=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)} \\ $$$${r}×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{n}+\mathrm{1}−\mathrm{1}} {xcos}^{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dx} \\ $$$$={r}×\frac{\lceil\left(\mathrm{2}{n}+\mathrm{1}\right)\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\lceil\left(\mathrm{2}{n}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$${i}\:{have}\:{tried}\:{to}\:{solve}..{others}\:{pls}\:{check}… \\ $$