calculate-A-n-0-1-cos-narcosx-dx-with-n-integr-natural-

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

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}