let A_p =∫_0 ^π x^p cos(nx)dx 1) calculate A_0 ,A_1 ,A_2 2)determine a relation of recurrence between A


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Question Number 67795 by mathmax by abdo last updated on 31/Aug/19

$l e t A_{p} = \int_{0}^{π} x^{p} c o s (n x) d x$ $1) c a l c u l a t e A_{0}, A_{1}, A_{2}$ $2) d e t e r m i n e a r e l a t i o n o f r e c u r r e n c e b e t w e e n A_{p}$
Commented by mathmax by abdo last updated on 03/Sep/19
$1) A_0 =∫_0 ^π cos(nx)dx =[(1/n)sin(nx)]_0 ^π =0 A_1 =∫_0 ^π x cos(nx)dx by parts A_1 =[(x/n)sin(nx)]_0 ^(π ) −∫_0 ^π (1/n)sin(nx)dx =−(1/n) ∫_0 ^π sin(nx)dx =−(1/n)[−(1/n)cos(nx)]_0 ^π =(1/n^2 ){ (−1)^n −1} A_2 =∫_0 ^π x^2 cos(nx)dx =_(by parts) [(x^2 /n)sin(nx)]_0 ^π −∫_0 ^π ((2x)/n)sin(nx)dx =−(1/n)∫_0 ^π xsin(nx)dx =−(1/n){ [−(x/n)cos(nx)]_0 ^π −∫_0 ^π −(1/n)cos(nx)dx} =−(1/n){−(1/n)(π(−1)^n ) +(1/n^2 )[sin(nx)]_0 ^π } =(π/n^2 )(−1)^n$
$1) A_{0} = \int_{0}^{π} c o s (n x) d x = {[\frac{1}{n} s i n (n x)]}_{0}^{π} = 0$ $A_{1} = \int_{0}^{π} x c o s (n x) d x b y p a r t s$ $A_{1} = {[\frac{x}{n} s i n (n x)]}_{0}^{π} - \int_{0}^{π} \frac{1}{n} s i n (n x) d x$ $= - \frac{1}{n} \int_{0}^{π} s i n (n x) d x = - \frac{1}{n} {[- \frac{1}{n} c o s (n x)]}_{0}^{π} = \frac{1}{n^{2}} {{(- 1)}^{n} - 1}$ $A_{2} = \int_{0}^{π} x^{2} c o s (n x) d x =_{b y p a r t s} {[\frac{x^{2}}{n} s i n (n x)]}_{0}^{π} - \int_{0}^{π} \frac{2 x}{n} s i n (n x) d x$ $= - \frac{1}{n} \int_{0}^{π} x s i n (n x) d x = - \frac{1}{n} {{[- \frac{x}{n} c o s (n x)]}_{0}^{π} - \int_{0}^{π} - \frac{1}{n} c o s (n x) d x}$ $= - \frac{1}{n} {- \frac{1}{n} (π {(- 1)}^{n}) + \frac{1}{n^{2}} {[s i n (n x)]}_{0}^{π}}$ $= \frac{π}{n^{2}} {(- 1)}^{n}$
Commented by mathmax by abdo last updated on 04/Sep/19

$2) b y p a r t s u^{^{'}} = x^{p} a n d v = c o s (n x) \Rightarrow$ $A_{p} = {[\frac{x^{p + 1}}{p + 1} c o s (n x)]}_{0}^{π} - \int_{0}^{π} \frac{x^{p + 1}}{p + 1} (- n s i n (n x)) d x$ $= \frac{π^{p + 1}}{p + 1} {(- 1)}^{n} + \frac{n}{p + 1} \int_{0}^{π} x^{p + 1} s i n (n x) d x a g a i n b y p a r t s$ $u^{^{'}} = x^{p + 1} a n d v = s i n (n x) \Rightarrow$ $\int_{0}^{π} x^{p + 1} s i n (n x) d x = {[\frac{x^{p + 2}}{p + 2} s i n (n x)]}_{0}^{π} - \int_{0}^{π} \frac{x^{p + 2}}{p + 2} \frac{1}{n} c o s (n x) d x$ $= - \frac{1}{n (p + 2)} \int_{0}^{π} x^{p + 2} c o s (n x) d x = - \frac{1}{n (p + 2)} A_{p + 2} \Rightarrow$ $A_{p} = \frac{π^{p + 1}}{p + 1} {(- 1)}^{n} + \frac{n}{p + 1} (- \frac{1}{n (p + 2)}) A_{p + 2} \Rightarrow$ $A_{p} = \frac{π^{p + 1}}{p + 1} {(- 1)}^{n} - \frac{1}{(p + 1) (p + 2)} A_{p + 2} \Rightarrow$ $\frac{1}{(p + 1) (p + 2)} A_{p + 2} = \frac{π^{p + 1}}{p + 1} {(- 1)}^{n} - A_{p} \Rightarrow$ $A_{p + 2} = (p + 1) (p + 2) \frac{π^{p + 1}}{p + 1} {(- 1)}^{n} - (p + 1) (p + 2) A_{p} \Rightarrow$ $A_{p + 2} = (p + 2) {(- 1)}^{n} π^{p + 1} - (p + 1) (p + 2) A_{p}$
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