calculate-I-n-0-1-sin-narcsinx-dx-

Question Number 84574 by msup trace by abdo last updated on 14/Mar/20

c a l c u l a t e I_{n} = \int_{0}^{1} s i n (n a r c s i n x) d x

Commented by mathmax by abdo last updated on 15/Mar/20

c h a n g e m e n t a r c s i n x = t g i v e x = s i n t \Rightarrow

I_{n} = \int_{0}^{\frac{π}{2}} s i n (n t) c o s t d t w e h a v e

s i n (n t) c o s t = c o s (\frac{π}{2} - n t) c o s t = \frac{1}{2} {c o s (\frac{π}{2} - n t + t) + c o s (\frac{π}{2} - n t - t)}

= \frac{1}{2} {c o s (\frac{π}{2} - (n - 1) t) + 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

I_{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) t) 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}} (n \neq 1)

= - \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}

= - \frac{1}{2 (n - 1)} {s i n (\frac{n π}{2}) - 1} + \frac{1}{2 (n + 1)} {s i n (\frac{n π}{2}) + 1}

= s i n (\frac{n π}{2}) (\frac{1}{2 (n + 1)} - \frac{1}{2 (n - 1)}) + \frac{1}{2 (n - 1)} + \frac{1}{2 (n + 1)}

= \frac{1}{2} s i n (\frac{n π}{2}) (\frac{- 2}{n^{2} - 1}) + \frac{1}{2} (\frac{2 n}{n^{2} - 1})

= - \frac{s i n (\frac{n π}{2})}{n^{2} - 1} + \frac{n}{n^{2} - 1} \Rightarrow I_{n} = \frac{1}{n^{2} - 1} (n - s i n (\frac{n π}{2}))

I_{1} = \int_{0}^{\frac{π}{2}} s i n t c o s t d t = \frac{1}{2} \int_{0}^{\frac{π}{2}} s i n (2 t) d t = - \frac{1}{4} {[c o s (2 t)]}_{0}^{\frac{π}{2}}

= - \frac{1}{4} (- 2) = \frac{1}{2}