if-I-n-x-pi-2-xcos-n-xdx-where-n-1-show-that-I-n-n-n-1-I-n-2-1-n-2-and-then-evaluate-x-pi-2-xcos-8-xdx-

Question Number 115014 by mathdave last updated on 23/Sep/20

i f I_{n} = \int_{x}^{\frac{π}{2}} x \cos^{n} x d x, w h e r e n ≻ 1 s h o w

t h a t I_{n} = \frac{n (n - 1) I_{n - 2} - 1}{n^{2}} a n d t h e n

e v a l u a t e \int_{x}^{\frac{π}{2}} x \cos^{8} x d x

Answered by Olaf last updated on 23/Sep/20

I_{n} = \int_{0}^{\frac{π}{2}} x \cos^{n} x d x

I_{n} - I_{n + 2} = \int_{0}^{\frac{π}{2}} x \cos^{n} x (1 - \cos^{2} x) d x

I_{n} - I_{n + 2} = - \int_{0}^{\frac{π}{2}} (x \sin x) (- \sin x \cos^{n} x) d x

I_{n} - I_{n + 2} = - {[x \sin x \frac{\cos^{n + 1} x}{n + 1}]}_{0}^{\frac{π}{2}}

+ \int_{0}^{\frac{π}{2}} (\sin x + x \cos x) \frac{\cos^{n + 1} x}{n + 1} d x

I_{n} - I_{n + 2} = - \int_{0}^{\frac{π}{2}} (- \sin x \frac{\cos^{n + 1} x}{n + 1}) d x

+ \frac{1}{n + 1} I_{n + 2}

I_{n} - I_{n + 2} = - {[\frac{\cos^{n + 2} x}{(n + 1) (n + 2)}]}_{0}^{\frac{π}{2}} + \frac{1}{n + 1} I_{n + 2}

I_{n} = \frac{1}{(n + 1) (n + 2)} + I_{n + 2} + \frac{1}{n + 1} I_{n + 2}

I_{n} = \frac{n + 2}{n + 1} I_{n + 2} + \frac{1}{(n + 1) (n + 2)}

I_{n + 2} = \frac{n + 1}{n + 2} I_{n} - \frac{1}{{(n + 2)}^{2}} (1)

I_{n} = \frac{n (n - 1) I_{n - 2} - 1}{n^{2}}

I_{0} = \int_{0}^{\frac{π}{2}} x d x = {[\frac{x^{2}}{2}]}_{0}^{\frac{π}{2}} = \frac{π^{2}}{8}

With (1) : I_{2} = \frac{1}{2} (\frac{π^{2}}{8}) - \frac{1}{4} = \frac{π^{2}}{16} - \frac{1}{4}

I_{4} = \frac{3}{4} (\frac{π^{2}}{16} - \frac{1}{4}) - \frac{1}{16} = \frac{3 π^{2}}{64} - \frac{1}{4}

I_{6} = \frac{5}{6} (\frac{3 π^{2}}{64} - \frac{1}{4}) - \frac{1}{36} = \frac{5 π^{2}}{128} - \frac{17}{72}

I_{8} = \frac{7}{8} (\frac{5 π^{2}}{128} - \frac{17}{72}) - \frac{1}{64} = \frac{35 π^{2}}{1024} - \frac{2}{9}

I_{8} = \int_{0}^{\frac{π}{2}} x \cos^{8} x d x = \frac{35 π^{2}}{1024} - \frac{2}{9}

Commented by Tawa11 last updated on 06/Sep/21

great sir

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