Prove-that-I-n-0-pi-2-dt-1-tant-n-does-not-depend-of-the-term-n-deduces-that-0-dx-x-2035-1-x-2-1-pi-4-

Question Number 65805 by ~ À ® @ 237 ~ last updated on 04/Aug/19

P r o v e t h a t I_{n} = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + {(t a n t)}^{n}} d o e s n o t d e p e n d o f t h e t e r m n

d e d u c e s t h a t

\int_{0}^{\infty} \frac{d x}{(x^{2035} + 1) (x^{2} + 1)} = \frac{π}{4}

Commented by mathmax by abdo last updated on 04/Aug/19

I_{n} = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + {(t a n t)}^{n}} c h a n g e m e n t t = \frac{π}{2} - x g i v e

I_{n} = - \int_{0}^{\frac{π}{2}} \frac{- d x}{1 + \frac{1}{{t a n}^{n} x}} = \int_{0}^{\frac{π}{2}} \frac{{t a n}^{n} x}{1 + {t a n}^{n} x} d x = \int_{0}^{\frac{π}{2}} \frac{1 + {t a n}^{n} x - 1}{1 + {t a n}^{n} x} d x

= \frac{π}{2} - I_{n} \Rightarrow 2 I_{n} = \frac{π}{2} \Rightarrow I_{n} = \frac{π}{4} \forall n

c h a n g e m e n t t a n t = x g i v e I_{n} = \int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (1 + x^{n})} \Rightarrow

\int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (1 + x^{n})} = \frac{π}{4} l e t t a k e n = 2035 \Rightarrow

\int_{0}^{\infty} \frac{d x}{(x^{2035} + 1) (1 + x^{2})} = \frac{π}{4}

Answered by Tanmay chaudhury last updated on 04/Aug/19

I (n) = \int_{0}^{\frac{π}{2}} \frac{{c o s}^{n} t}{{s i n}^{n} t + {c o s}^{n} t} d t \dots . . (1)

= \int_{0}^{\frac{π}{2}} \frac{{c o s}^{n} (\frac{π}{2} - t)}{{s i n}^{n} (\frac{π}{2} - t) + {c o s}^{n} (\frac{π}{2} - t)} d t

= \int_{0}^{\frac{π}{2}} \frac{{s i n}^{n} t}{{c o s}^{n} t + {s i n}^{n} t} d t \dots . . (2)

2 I (n) = \int_{0}^{\frac{π}{2}} d t \dots . . (b y a d d i n g (1) a n d 2)

I (n) = \frac{1}{2} \times \frac{π}{2} = \frac{π}{4}

Answered by Tanmay chaudhury last updated on 04/Aug/19

x = t a n a d x = {s e c}^{2} a d a

\int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} a d a}{{t a n}^{2035} a + 1} \times \frac{1}{{s e c}^{2} a} d a

= \frac{π}{4}