Question Number 41579 by Dawajan Nikmal last updated on 09/Aug/18
$$\mathrm{If}\:\:{u}_{\mathrm{10}} =\:\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}{x}^{\mathrm{10}} \:\mathrm{sin}\:{x}\:{dx},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$${u}_{\mathrm{10}} +\mathrm{90}\:{u}_{\mathrm{8}} \:\:\mathrm{is} \\ $$
Commented by maxmathsup by imad last updated on 11/Aug/18
![let u_n = ∫_0 ^(π/2) x^n sinx dx let find u_n by recurrence by parts we have u_n =[(1/(n+1)) x^(n+1) sinx]_0 ^(π/2) −∫_0 ^(π/2) (1/(n+1))x^(n+1) cosx dx =(π^(n+1) /((n+1)2^(n+1) )) −(1/(n+1)) ∫_0 ^(π/2) x^(n+1) cosx dx =(π^(n+1) /((n+1)2^(n+1) )) −(1/((n+1))){ [(1/(n+2))x^(n+2) cosx]_0 ^(π/2) +∫_0 ^(π/2) (1/((n+2))) x^(n+2) sinxdx} =(π^(n+1) /((n+1)2^(n+1) )) −(1/((n+1)(n+2))) u_(n+2) ⇒ u_8 =(π^9 /(9 .2^9 )) −(1/(9.10)) u_(10) ⇒ u_(10) +90 u_8 = u_(10) +90 (π^9 /(9.2^9 )) −u_(10) ⇒ u_(10) +90 u_8 = ((10×π^9 )/2^9 ) =10((π/2))^9 .](https://www.tinkutara.com/question/Q41670.png)
$${let}\:{u}_{{n}} =\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{x}^{{n}} {sinx}\:{dx}\:{let}\:{find}\:{u}_{{n}} \:{by}\:{recurrence}\:\:{by}\:{parts}\:{we}\:{have} \\ $$$${u}_{{n}} \:=\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}\:{x}^{{n}+\mathrm{1}} \:{sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:{cosx}\:{dx} \\ $$$$=\frac{\pi^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)\mathrm{2}^{{n}+\mathrm{1}} }\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{x}^{{n}+\mathrm{1}} \:{cosx}\:{dx} \\ $$$$=\frac{\pi^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)\mathrm{2}^{{n}+\mathrm{1}} }\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)}\left\{\:\:\left[\frac{\mathrm{1}}{{n}+\mathrm{2}}{x}^{{n}+\mathrm{2}} \:{cosx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)}\:{x}^{{n}+\mathrm{2}} \:{sinxdx}\right\} \\ $$$$=\frac{\pi^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)\mathrm{2}^{{n}+\mathrm{1}} }\:−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\:{u}_{{n}+\mathrm{2}} \:\:\Rightarrow \\ $$$${u}_{\mathrm{8}} =\frac{\pi^{\mathrm{9}} }{\mathrm{9}\:.\mathrm{2}^{\mathrm{9}} }\:−\frac{\mathrm{1}}{\mathrm{9}.\mathrm{10}}\:{u}_{\mathrm{10}} \:\:\Rightarrow\:{u}_{\mathrm{10}} \:+\mathrm{90}\:{u}_{\mathrm{8}} \:=\:{u}_{\mathrm{10}} \:\:+\mathrm{90}\:\frac{\pi^{\mathrm{9}} }{\mathrm{9}.\mathrm{2}^{\mathrm{9}} }\:−{u}_{\mathrm{10}} \:\Rightarrow \\ $$$${u}_{\mathrm{10}} \:+\mathrm{90}\:{u}_{\mathrm{8}} =\:\frac{\mathrm{10}×\pi^{\mathrm{9}} }{\mathrm{2}^{\mathrm{9}} }\:=\mathrm{10}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{9}} \:. \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18
![∫x^(10) sinxdx =x^(10) .(−cosx)−∫10x^9 .(−cosx)dx =−x^(10) cosx+10∫x^9 cosxdx =−x^(10) cosx+10[(x^9 sinx−9∫x^8 sinxdx] =−x^(10) cosx+10x^9 sinx−90∫x^8 sinxdx so ∫_0 ^(Π/2) x^(10) sinxdx =∣−x^(10) cosx+10x^9 sinx∣_0 ^(Π/2) +(−90)∫_0 ^(Π/2) x^8 sinx u_(10) +90u_8 ={−((Π/2))^(10) cos(Π/2)+10.((Π/2))^9 sin(Π/2)} u_(10) +90u_8 =10.((Π/2))^9](https://www.tinkutara.com/question/Q41583.png)
$$\int{x}^{\mathrm{10}} {sinxdx} \\ $$$$={x}^{\mathrm{10}} .\left(−{cosx}\right)−\int\mathrm{10}{x}^{\mathrm{9}} .\left(−{cosx}\right){dx} \\ $$$$=−{x}^{\mathrm{10}} {cosx}+\mathrm{10}\int{x}^{\mathrm{9}} {cosxdx} \\ $$$$=−{x}^{\mathrm{10}} {cosx}+\mathrm{10}\left[\left({x}^{\mathrm{9}} {sinx}−\mathrm{9}\int{x}^{\mathrm{8}} {sinxdx}\right]\right. \\ $$$$=−{x}^{\mathrm{10}} {cosx}+\mathrm{10}{x}^{\mathrm{9}} {sinx}−\mathrm{90}\int{x}^{\mathrm{8}} {sinxdx} \\ $$$${so}\:\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {x}^{\mathrm{10}} {sinxdx} \\ $$$$=\mid−{x}^{\mathrm{10}} {cosx}+\mathrm{10}{x}^{\mathrm{9}} {sinx}\mid_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \:+\left(−\mathrm{90}\right)\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {x}^{\mathrm{8}} {sinx} \\ $$$${u}_{\mathrm{10}} +\mathrm{90}{u}_{\mathrm{8}} =\left\{−\left(\frac{\Pi}{\mathrm{2}}\right)^{\mathrm{10}} {cos}\frac{\Pi}{\mathrm{2}}+\mathrm{10}.\left(\frac{\Pi}{\mathrm{2}}\right)^{\mathrm{9}} {sin}\frac{\Pi}{\mathrm{2}}\right\} \\ $$$${u}_{\mathrm{10}} +\mathrm{90}{u}_{\mathrm{8}} =\mathrm{10}.\left(\frac{\Pi}{\mathrm{2}}\right)^{\mathrm{9}} \\ $$