find-I-n-dx-sin-n-x-with-n-integr-natural-

Question Number 59732 by maxmathsup by imad last updated on 14/May/19

f i n d I_{n} = \int \frac{d x}{{s i n}^{n} x} w i t h n i n t e g r n a t u r a l .

Answered by tanmay last updated on 14/May/19

I_{n} = \int {c o s e c}^{n} x d x

l e t n = e v e n n u m b e r

\int {c o s e c}^{n - 2} x {c o z e c}^{2} x d x

\int {({c o s e c}^{2} x)}^{\frac{n - 2}{2}} {c o s e c}^{2} x d x

\int {(1 + {c o t}^{2} x)}^{\frac{n - 2}{2}} {c o s e c}^{2} x d x

a = c o t x

d a = - {c o s e c}^{2} x d x

\int {(1 + a^{2})}^{\frac{n - 2}{2}} \times - d a

= (- 1) \int [1 + \frac{n - 2}{2} C_{1} a^{2} + \frac{n - 2}{2} C_{2} a^{4} + \dots + {(a^{2})}^{\frac{n - 2}{2}}] d a

= (- 1) [a + \frac{n - 2}{2} C_{1} \times \frac{a^{3}}{3} + \frac{n - 2}{2} C_{2} \times \frac{a^{5}}{5} + . . + \frac{a^{n - 1}}{n - 1}]

n o w p l s p u t c o t x = a]

w h e n n = o d d

i a m t r y i n g \dots

Commented by Tawa1 last updated on 14/May/19

God bless you sir

Answered by MJS last updated on 14/May/19

trying Weierstrass

x = 2 \arctan t

\sin x = \frac{2 t}{t^{2} + 1}; d x = \frac{2 d t}{t^{2} + 1}

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

= \frac{1}{2^{n - 1}} \int (\underset{k = 1}{\sum^{n}} (\begin{matrix} n - 1 \\ k - 1 \end{matrix}) \frac{t^{2 k - 2}}{t^{n}}) d t =

= \frac{(n - 1)!}{2^{n - 1}} \int (\frac{1}{t^{2}} \underset{k = 1}{\sum^{n}} \frac{t^{2 k - n}}{(k - 1)! (n - k)!}) d t

I_{1} = \int \frac{1}{t} d t

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

I_{3} = \frac{1}{4} \int (\frac{1}{t^{3}} + \frac{2}{t} + t) d t

I_{4} = \frac{1}{8} \int (\frac{1}{t^{4}} + \frac{3}{t^{2}} + 3 + t^{2}) d t

I_{5} = \frac{1}{16} \int (\frac{1}{t^{5}} + \frac{4}{t^{3}} + \frac{6}{t} + 4 t + t^{3}) d t

I_{6} = \frac{1}{32} \int (\frac{1}{t^{6}} + \frac{5}{t^{4}} + \frac{10}{t^{2}} + 10 + 5 t^{2} + t^{4}) d t

\dots

Commented by maxmathsup by imad last updated on 14/May/19

t h a n k y o u s i r .