Integration Questions

Question Number 172741 by Kalebwizeman last updated on 30/Jun/22

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}{Sin}\overset{\mathrm{100}} {\:}{xdx} \\$$

Answered by aleks041103 last updated on 01/Jul/22

$${I}_{{n}} =\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {sin}^{{n}} \left({x}\right)\:{dx} \\$$$${sin}^{{n}} {x}=\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right){sin}^{{n}−\mathrm{2}} {x} \\$$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}{I}_{{n}} =\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {sin}^{{n}−\mathrm{2}} {x}\:{dx}−\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {cos}^{\mathrm{2}} {x}\:{sin}^{{n}−\mathrm{2}} {x}\:{dx}= \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}{I}_{{n}−\mathrm{2}} −\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left({cosx}\right)\left({cosx}\:{sin}^{{n}−\mathrm{2}} {x}\:{dx}\right)= \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}{I}_{{n}−\mathrm{2}} −\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left({cosx}\right)\left(\left({sinx}\right)^{{n}−\mathrm{2}} {d}\left({sinx}\right)\right)= \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}{I}_{{n}−\mathrm{2}} −\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left({cosx}\right){d}\left(\frac{{sin}^{{n}−\mathrm{1}} {x}}{{n}−\mathrm{1}}\right)= \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}{I}_{{n}−\mathrm{2}} −\frac{\mathrm{1}}{{n}−\mathrm{1}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left({cosx}\right){d}\left({sin}^{{n}−\mathrm{1}} {x}\right) \\$$$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left({cosx}\right){d}\left({sin}^{{n}−\mathrm{1}} {x}\right)= \\$$$$=\left[{cosxsin}^{{n}−\mathrm{1}} {x}\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} −\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left({sin}^{{n}−\mathrm{1}} {x}\right){d}\left({cosx}\right)= \\$$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {sin}^{{n}} {x}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}{I}_{{n}} \\$$$$\Rightarrow{I}_{{n}} ={I}_{{n}−\mathrm{2}} −\frac{\mathrm{1}}{{n}−\mathrm{1}}{I}_{{n}} \\$$$$\Rightarrow\left(\mathrm{1}+\frac{\mathrm{1}}{{n}−\mathrm{1}}\right){I}_{{n}} ={I}_{{n}−\mathrm{2}} \\$$$$\Rightarrow{I}_{{n}} =\frac{{n}−\mathrm{1}}{{n}}{I}_{{n}−\mathrm{2}} \\$$$$\Rightarrow{I}_{\mathrm{2}{m}} =\frac{\mathrm{2}{m}−\mathrm{1}}{\mathrm{2}{m}}{I}_{\mathrm{2}\left({m}−\mathrm{1}\right)} \\$$$$\Rightarrow{I}_{\mathrm{2}{m}} =\left(\underset{{s}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\frac{\mathrm{2}\left({m}−{s}\right)−\mathrm{1}}{\mathrm{2}\left({m}−{s}\right)}\right){I}_{\mathrm{2}\left({m}−{k}\right)} \\$$$${k}={m} \\$$$$\Rightarrow{I}_{\mathrm{2}{m}} =\left(\underset{{s}=\mathrm{0}} {\overset{{m}−\mathrm{1}} {\prod}}\frac{\mathrm{2}\left({m}−{s}\right)−\mathrm{1}}{\mathrm{2}\left({m}−{s}\right)}\right){I}_{\mathrm{0}} \\$$$${I}_{\mathrm{0}} =\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {sin}^{\mathrm{0}} {x}\:{dx}=\pi \\$$$$\Rightarrow{I}_{\mathrm{2}{m}} =\pi\underset{{m}−{s}={s}'=\mathrm{1}} {\overset{{m}} {\prod}}\frac{\mathrm{2}{s}'−\mathrm{1}}{\mathrm{2}{s}'} \\$$$${I}_{\mathrm{2}{m}} =\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\:...\:.\left(\mathrm{2}{m}−\mathrm{1}\right)}{\mathrm{2}.\mathrm{4}.\mathrm{6}.\mathrm{8}.\:...\:.\left(\mathrm{2}{m}\right)}\:\pi \\$$$$\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\:...\:.\left(\mathrm{2}{m}−\mathrm{1}\right)}{\mathrm{2}.\mathrm{4}.\mathrm{6}.\mathrm{8}.\:...\:.\left(\mathrm{2}{m}\right)}\:=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\:...\:.\left(\mathrm{2}{m}−\mathrm{1}\right)\left(\mathrm{2}{m}\right)}{\left(\mathrm{2}.\mathrm{4}.\mathrm{6}.\mathrm{8}.\:...\:.\left(\mathrm{2}{m}\right)\right)^{\mathrm{2}} }=\frac{\left(\mathrm{2}{m}\right)!}{\left(\mathrm{2}^{{m}} {m}!\right)^{\mathrm{2}} } \\$$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} {sin}^{\mathrm{2}{m}} {x}\:{dx}=\frac{\left(\mathrm{2}{m}\right)!\:\pi}{\mathrm{4}^{{m}} \left({m}!\right)^{\mathrm{2}} } \\$$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} {sin}^{\mathrm{100}} {x}\:{dx}=\frac{\mathrm{100}!\pi}{\mathrm{4}^{\mathrm{100}} \left(\mathrm{50}!\right)^{\mathrm{2}} } \\$$

Commented by Tawa11 last updated on 01/Jul/22

$$\mathrm{Great}\:\mathrm{sir} \\$$

Commented by Kalebwizeman last updated on 02/Jul/22

$${thank}\:{you}\:{so}\:{much}\:{sir} \\$$