cos-n-x-sinx-dx-

Question Number 148077 by mathdanisur last updated on 25/Jul/21

\int \frac{{c o s}^{n} x}{s i n x} d x = ?

Answered by Olaf_Thorendsen last updated on 26/Jul/21

I_{n} (x) = \int \frac{\cos^{n} x}{\sin x} d x

∙ I_{0} (x) = \int \frac{d x}{\sin x} d x

Let t = \tan \frac{x}{2}

I_{0} (x) = \int \frac{1}{\frac{2 t}{1 + t^{2}}} . \frac{2 d t}{1 + t^{2}}

I_{0} (x) = \ln ∣ t ∣ + C = \ln ∣ \tan \frac{x}{2} ∣ + C

∙ I_{1} (x) = \int \frac{\cos x}{\sin x} d x = \ln ∣ \sin x ∣ + C

∙ k ⩾ 2

I_{k} (x) - I_{k - 2} (x) = \int \frac{\cos^{k - 2} x (\cos^{2} x - 1)}{\sin x} d x

I_{k} (x) - I_{k - 2} (x) = - \int \sin x . \cos^{k - 2} x d x

I_{k} (x) - I_{k - 2} (x) = \frac{1}{k - 1} \cos^{k - 1} x

∙ If k is even, k = 2 p

\underset{p = 1}{\sum^{n}} I_{2 p} (x) - \underset{p = 1}{\sum^{n}} I_{2 p - 2} (x) = \underset{p = 1}{\sum^{n}} \frac{1}{2 p - 1} \cos^{2 p - 1} x

I_{2 n} (x) - I_{0} (x) = \underset{p = 1}{\sum^{n}} \frac{\cos^{2 p - 1} x}{2 p - 1}

I_{2 n} (x) = \ln ∣ \tan \frac{x}{2} ∣ + \underset{p = 1}{\sum^{n}} \frac{\cos^{2 p - 1} x}{2 p - 1} + C

∙ If k is odd, k = 2 p + 1

\underset{p = 1}{\sum^{n}} I_{2 p + 1} (x) - \underset{p = 1}{\sum^{n}} I_{2 p - 1} (x) = \underset{p = 1}{\sum^{n}} \frac{1}{2 p} \cos^{2 p} x

I_{2 n + 1} (x) - I_{1} (x) = \frac{1}{2} \underset{p = 1}{\sum^{n}} \frac{\cos^{2 p} x}{p}

I_{2 n + 1} (x) = \ln ∣ \sin x ∣ + \frac{1}{2} \underset{p = 1}{\sum^{n}} \frac{\cos^{2 p} x}{p} + C

Commented by mathdanisur last updated on 25/Jul/21

T h a n k y o u S i r, a n s w e r . ?

Commented by puissant last updated on 26/Jul/21

p r o f d e s o l e m a i s I_{1} = l n ∣ s i n (x) ∣ + C

c a r {(l n (u))}^{'} = \frac{u^{'}}{u} a v e c u = s i n (x) . .

Commented by Olaf_Thorendsen last updated on 26/Jul/21

Bien vu!

(mais je ne suis pas prof) .

Commented by puissant last updated on 26/Jul/21

d^{'} a c c o r d m o n s i e u r o l a f . .

Answered by mathmax by abdo last updated on 25/Jul/21

A_{n} = \int \frac{\cos^{n} x}{sinx} dx \Rightarrow A_{2 n} = \int \frac{\cos^{2 n} x}{sinx} dx

= \int \frac{{(1 - \sin^{2} x)}^{n}}{sinx} dx = {(- 1)}^{n} \int \frac{{(\sin^{2} x - 1)}^{n}}{sinx} dx

= {(- 1)}^{n} \int \frac{\sum_{k = 0}^{n} C_{n}^{k} \sin^{2 k} x {(- 1)}^{n - k}}{sinx} dx

= \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} \int \sin^{2 k - 1} x dx

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

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

= \int \frac{dx}{sinx} + \sum_{k = 0}^{n} \frac{{(- 1)}^{k}}{2^{2 k - 1}} C_{n}^{k} (\sum_{p = 0}^{2 k - 1} {(- 1)}^{2 k - 1 - p} \int e^{ipx} e^{- i (2 k - 1) x + ipxdx}

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

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

rest to find A_{2 n + 1} \dots . be continued \dots .

Commented by mathdanisur last updated on 25/Jul/21

T h a n k y o u S i r, a n s w e r . ?

Commented by mathmax by abdo last updated on 25/Jul/21

A_{2 n} = \int \frac{dx}{sinx} + \sum_{k = 1}^{n} (\dots .)