cos-2020-x-dx-

Question Number 85872 by Wepa last updated on 25/Mar/20

\int \cos^{2020} x dx = ?

Commented by mr W last updated on 25/Mar/20

I_{n} = \int \cos^{n} x d x

= \int \cos^{n - 1} x d (\sin x)

= \sin x \cos^{n - 1} x + (n - 1) \int \sin^{2} x \cos^{n - 2} x d x

= \sin x \cos^{n - 1} x + (n - 1) \int (1 - \cos^{2} x) \cos^{n - 2} x d x

= \sin x \cos^{n - 1} x + (n - 1) \int \cos^{n - 2} x d x - (n - 1) \int \cos^{n} x d x

= \sin x \cos^{n - 1} x + (n - 1) I_{n - 2} - (n - 1) I_{n}

{n I}_{n} = \sin x \cos^{n - 1} x + (n - 1) I_{n - 2}

\Rightarrow I_{n} = \frac{1}{n} \sin x \cos^{n - 1} x + \frac{n - 1}{n} I_{n - 2}

Answered by mind is power last updated on 26/Mar/20

\int {c o s}^{n} (x) d x

- \int \frac{u^{m}}{\sqrt{1 - u^{2}}} d u

\frac{1}{\sqrt{1 - u^{2}}} = \sum_{n ⩾ 0} \frac{(2 n)!}{2^{2 n} {(n!)}^{2}} u^{2 n + m}

= - \int \frac{(2 n)! u^{2 n + m}}{2^{2 n} {(n!)}^{2}} d u

= - \sum_{n ⩾ 0} \frac{(2 n)! u^{2 n + m + 1}}{2^{2 n} {(n!)}^{2} (2 n + m + 1)} = - x^{m + 1} \sum_{n ⩾ 0} \frac{\underset{k = 0}{\prod^{n - 1}} (k + \frac{1}{2})}{(2 n + m + 1)} . {(\frac{u^{2}}{2})}^{n}

= - x^{m + 1} . \frac{2}{m + 1} \sum_{k ⩾ 0} \frac{\underset{k = 0}{\prod^{n - 1}} (\frac{1}{2} + k) . \underset{j = 0}{\prod^{n - 1}} (\frac{m + 1}{2} + j)}{2 \underset{j = 0}{\prod^{n - 1}} (j + \frac{m + 3}{2})} . {(\frac{u^{2}}{2})}^{n} . \frac{1}{n!}

= \frac{- 2 u^{m + 1}}{m + 1} ._{2} F_{1} (\frac{1}{2}, \frac{m + 1}{2}; \frac{m + 3}{2}; \frac{u^{2}}{2}) + c

\frac{- 2 {c o s}^{m + 1} (x)}{m + 1}_{2} F_{1} (\frac{1}{2}; \frac{m + 1}{2}; \frac{m + 3}{2}; \frac{{c o s}^{2} (x)}{2}) + c