If u_(10) = ∫_( 0) ^(π/2) x^(10) sin x dx, then the value of u_(10) +90 u


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Question Number 41579 by Dawajan Nikmal last updated on 09/Aug/18

$If u_{10} = \underset{0}{\int^{π / 2}} x^{10} \sin x d x, then the value of$ $u_{10} + 90 u_{8} is$
Commented by maxmathsup by imad last updated on 11/Aug/18
$let u_n = ∫_0 ^(π/2) x^n sinx dx let find u_n by recurrence by parts we have u_n =[(1/(n+1)) x^(n+1) sinx]_0 ^(π/2) −∫_0 ^(π/2) (1/(n+1))x^(n+1) cosx dx =(π^(n+1) /((n+1)2^(n+1) )) −(1/(n+1)) ∫_0 ^(π/2) x^(n+1) cosx dx =(π^(n+1) /((n+1)2^(n+1) )) −(1/((n+1))){ [(1/(n+2))x^(n+2) cosx]_0 ^(π/2) +∫_0 ^(π/2) (1/((n+2))) x^(n+2) sinxdx} =(π^(n+1) /((n+1)2^(n+1) )) −(1/((n+1)(n+2))) u_(n+2) ⇒ u_8 =(π^9 /(9 .2^9 )) −(1/(9.10)) u_(10) ⇒ u_(10) +90 u_8 = u_(10) +90 (π^9 /(9.2^9 )) −u_(10) ⇒ u_(10) +90 u_8 = ((10×π^9 )/2^9 ) =10((π/2))^9 .$
$l e t u_{n} = \int_{0}^{\frac{π}{2}} x^{n} s i n x d x l e t f i n d u_{n} b y r e c u r r e n c e b y p a r t s w e h a v e$ $u_{n} = {[\frac{1}{n + 1} x^{n + 1} s i n x]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{1}{n + 1} x^{n + 1} c o s x d x$ $= \frac{π^{n + 1}}{(n + 1) 2^{n + 1}} - \frac{1}{n + 1} \int_{0}^{\frac{π}{2}} x^{n + 1} c o s x d x$ $= \frac{π^{n + 1}}{(n + 1) 2^{n + 1}} - \frac{1}{(n + 1)} {{[\frac{1}{n + 2} x^{n + 2} c o s x]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} \frac{1}{(n + 2)} x^{n + 2} s i n x d x}$ $= \frac{π^{n + 1}}{(n + 1) 2^{n + 1}} - \frac{1}{(n + 1) (n + 2)} u_{n + 2} \Rightarrow$ $u_{8} = \frac{π^{9}}{9 . 2^{9}} - \frac{1}{9.10} u_{10} \Rightarrow u_{10} + 90 u_{8} = u_{10} + 90 \frac{π^{9}}{9 . 2^{9}} - u_{10} \Rightarrow$ $u_{10} + 90 u_{8} = \frac{10 \times π^{9}}{2^{9}} = 10 {(\frac{π}{2})}^{9} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18
	$∫x^(10) sinxdx =x^(10) .(−cosx)−∫10x^9 .(−cosx)dx =−x^(10) cosx+10∫x^9 cosxdx =−x^(10) cosx+10[(x^9 sinx−9∫x^8 sinxdx] =−x^(10) cosx+10x^9 sinx−90∫x^8 sinxdx so ∫_0 ^(Π/2) x^(10) sinxdx =∣−x^(10) cosx+10x^9 sinx∣_0 ^(Π/2) +(−90)∫_0 ^(Π/2) x^8 sinx u_(10) +90u_8 ={−((Π/2))^(10) cos(Π/2)+10.((Π/2))^9 sin(Π/2)} u_(10) +90u_8 =10.((Π/2))^9$
	$\int x^{10} s i n x d x$ $= x^{10} . (- c o s x) - \int 10 x^{9} . (- c o s x) d x$ $= - x^{10} c o s x + 10 \int x^{9} c o s x d x$ $= - x^{10} c o s x + 10 [(x^{9} s i n x - 9 \int x^{8} s i n x d x]$ $= - x^{10} c o s x + 10 x^{9} s i n x - 90 \int x^{8} s i n x d x$ $s o \int_{0}^{\frac{Π}{2}} x^{10} s i n x d x$ $=∣ - x^{10} c o s x + 10 x^{9} s i n x ∣_{0}^{\frac{Π}{2}} + (- 90) \int_{0}^{\frac{Π}{2}} x^{8} s i n x$ $u_{10} + 90 u_{8} = {- {(\frac{Π}{2})}^{10} c o s \frac{Π}{2} + 10 . {(\frac{Π}{2})}^{9} s i n \frac{Π}{2}}$ $u_{10} + 90 u_{8} = 10 . {(\frac{Π}{2})}^{9}$
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