calculate A_n =∫_0 ^1 cos(narcosx)dx with n integr natural


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Question Number 79646 by abdomathmax last updated on 27/Jan/20

$c a l c u l a t e A_{n} = \int_{0}^{1} c o s (n a r c o s x) d x$ $w i t h n i n t e g r n a t u r a l$
Commented by abdomathmax last updated on 12/Mar/20
$changement arcosx=t give x=cost ⇒ A_n =∫_(π/2) ^0 cos(nt)(−sint)dt =∫_0 ^(π/2) sint cos(nt)dt we have sint cos(nt) =cos((π/2)−t)cos(nt) =(1/2){ cos((π/2)−t+nt)+cos(nt−(π/2)+t)} =(1/2){ cos((n−1)t+(π/2))+cos((π/2)−(n+1)t) =(1/2){ −sin(n−1)t +sin(n+1)t} ⇒ A_n =(1/2)∫_0 ^(π/2) sin(n+1)t dt−(1/2)∫_0 ^(π/2) sin(n−1)dt =−(1/(2(n+1)))[cos(n+1)t]_0 ^(π/2) +(1/(2(n−1)))[cos(n−1)t]_0 ^(π/2) =−(1/(2(n+1))){ cos(n+1)(π/2)−1} +(1/(2(n−1))){ cos(n−1)(π/2)−1} (n≠1) A_1 =∫_0 ^1 xdx =(1/2)$
$c h a n g e m e n t a r c o s x = t g i v e x = c o s t \Rightarrow$ $A_{n} = \int_{\frac{π}{2}}^{0} c o s (n t) (- s i n t) d t$ $= \int_{0}^{\frac{π}{2}} s i n t c o s (n t) d t$ $w e h a v e s i n t c o s (n t)$ $= c o s (\frac{π}{2} - t) c o s (n t) = \frac{1}{2} {c o s (\frac{π}{2} - t + n t) + c o s (n t - \frac{π}{2} + t)}$ $= \frac{1}{2} {c o s ((n - 1) t + \frac{π}{2}) + c o s (\frac{π}{2} - (n + 1) t)$ $= \frac{1}{2} {- s i n (n - 1) t + s i n (n + 1) t} \Rightarrow$ $A_{n} = \frac{1}{2} \int_{0}^{\frac{π}{2}} s i n (n + 1) t d t - \frac{1}{2} \int_{0}^{\frac{π}{2}} s i n (n - 1) d t$ $= - \frac{1}{2 (n + 1)} {[c o s (n + 1) t]}_{0}^{\frac{π}{2}} + \frac{1}{2 (n - 1)} {[c o s (n - 1) t]}_{0}^{\frac{π}{2}}$ $= - \frac{1}{2 (n + 1)} {c o s (n + 1) \frac{π}{2} - 1}$ $+ \frac{1}{2 (n - 1)} {c o s (n - 1) \frac{π}{2} - 1} (n \neq 1)$ $A_{1} = \int_{0}^{1} x d x = \frac{1}{2}$
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