Question Number 84574 by msup trace by abdo last updated on 14/Mar/20
$${calculate}\:\:{I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{sin}\left({narcsinx}\right){dx} \\ $$
Commented by mathmax by abdo last updated on 15/Mar/20
![changement arcsinx=t give x=sint ⇒ I_n =∫_0 ^(π/2) sin(nt)cost dt we have sin(nt)cost =cos((π/2)−nt)cost =(1/2){cos((π/2)−nt +t)+cos((π/2)−nt−t)} =(1/2){cos((π/2)−(n−1)t)+cos((π/2)−(n+1)t)} =(1/2){sin(n−1)t+sin(n+1)t} ⇒ I_n =(1/2)∫_0 ^(π/2) sin((n−1)t) dt+(1/2)∫_0 ^(π/2) sin((n+1)t) dt =−(1/(2(n−1)))[cos(n−1)t]_0 ^(π/2) −(1/(2(n+1)))[cos(n+1)t]_0 ^(π/2) (n≠1) =−(1/(2(n−1))){cos(n−1)(π/2)−1}−(1/(2(n+1))){cos(n+1)(π/2)−1} =−(1/(2(n−1))){sin(((nπ)/2))−1}+(1/(2(n+1))){sin(((nπ)/2))+1} =sin(((nπ)/2))((1/(2(n+1)))−(1/(2(n−1))))+(1/(2(n−1))) +(1/(2(n+1))) =(1/2)sin(((nπ)/2))(((−2)/(n^2 −1)))+(1/2)(((2n)/(n^2 −1))) =−((sin(((nπ)/2)))/(n^2 −1)) +(n/(n^2 −1)) ⇒I_n =(1/(n^2 −1))(n−sin(((nπ)/2))) I_1 =∫_0 ^(π/2) sint cost dt =(1/2)∫_0 ^(π/2) sin(2t) dt =−(1/4)[cos(2t)]_0 ^(π/2) =−(1/4)(−2) =(1/2)](https://www.tinkutara.com/question/Q84764.png)
$${changement}\:{arcsinx}={t}\:{give}\:{x}={sint}\:\Rightarrow \\ $$$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left({nt}\right){cost}\:{dt}\:\:{we}\:{have} \\ $$$${sin}\left({nt}\right){cost}\:={cos}\left(\frac{\pi}{\mathrm{2}}−{nt}\right){cost}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\pi}{\mathrm{2}}−{nt}\:+{t}\right)+{cos}\left(\frac{\pi}{\mathrm{2}}−{nt}−{t}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{cos}\left(\frac{\pi}{\mathrm{2}}−\left({n}−\mathrm{1}\right){t}\right)+{cos}\left(\frac{\pi}{\mathrm{2}}−\left({n}+\mathrm{1}\right){t}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{sin}\left({n}−\mathrm{1}\right){t}+{sin}\left({n}+\mathrm{1}\right){t}\right\}\:\Rightarrow \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}\left(\left({n}−\mathrm{1}\right){t}\right)\:{dt}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left(\left({n}+\mathrm{1}\right){t}\right)\:{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}\left[{cos}\left({n}−\mathrm{1}\right){t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\left[{cos}\left({n}+\mathrm{1}\right){t}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\left({n}\neq\mathrm{1}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}\left\{{cos}\left({n}−\mathrm{1}\right)\frac{\pi}{\mathrm{2}}−\mathrm{1}\right\}−\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\left\{{cos}\left({n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}−\mathrm{1}\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}\left\{{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)−\mathrm{1}\right\}+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}\left\{{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)+\mathrm{1}\right\} \\ $$$$={sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}\right)+\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)\left(\frac{−\mathrm{2}}{{n}^{\mathrm{2}} −\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}{n}}{{n}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$=−\frac{{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)}{{n}^{\mathrm{2}} −\mathrm{1}}\:+\frac{{n}}{{n}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow{I}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} −\mathrm{1}}\left({n}−{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)\right) \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sint}\:{cost}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}\left(\mathrm{2}{t}\right)\:{dt}\:=−\frac{\mathrm{1}}{\mathrm{4}}\left[{cos}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left(−\mathrm{2}\right)\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$