cos-cosx-dx-

Question Number 143810 by Jamshidbek last updated on 18/Jun/21

\int \cos (cosx) dx = ?

Answered by mathmax by abdo last updated on 18/Jun/21

\int \cos (cosx) dx = \int \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} \cos^{2 n} x}{(2 n!} dx

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} \int \cos^{2 n} dx = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n)!} w_{n}

W_{n} = \int \cos^{2 n} dx (wallis generalise)

Answered by mathmax by abdo last updated on 19/Jun/21

\int \cos (cosx) dx = Re (\int e^{icosx} dx) but

\int e^{icosx} dx = \int \sum_{n = 0}^{\infty} \frac{i^{n} \cos^{n} x}{n!} dx = \sum_{n = 0}^{\infty} \frac{i^{n}}{n!} \int \cos^{n} dx and

\int \cos^{n} dx = \int {(\frac{e^{ix} + e^{- ix}}{2})}^{n} dx = \frac{1}{2^{n}} \int \sum_{k = 0}^{n} C_{n}^{k} e^{ikx} {(e^{- ix})}^{n - k}

= \frac{1}{2^{n}} \sum_{k = 0}^{n} C_{n}^{k} \int e^{ikx - i (n - k) x} dx

= \frac{1}{2^{n}} \sum_{k = 0}^{n} C_{n}^{k} \int e^{(2 ik - in) x} dx = \frac{1}{2^{n}} \sum_{k = 0}^{n} \frac{C_{n}^{k}}{2 ik - in} + k \Rightarrow

\int e^{icosx} dx = \sum_{n = 0}^{\infty} \frac{i^{n}}{n!} (\frac{1}{2^{n}} \sum_{k = 0}^{n} \frac{C_{n}^{k}}{i (2 k - n)}) + k

Commented by mathmax by abdo last updated on 20/Jun/21

\int e^{icosx} dx = \sum_{n = 0}^{\infty} \frac{i^{n}}{n! 2^{n}} \sum_{k = 0}^{n} \frac{C_{n}^{k}}{i (2 k - n)} e^{(2 k - n) ix} + K