calculate-1-x-cosx-i-1-n-1-tan-2-x-2-i-dx-

Question Number 65092 by mathmax by abdo last updated on 25/Jul/19

c a l c u l a t e \int \frac{1}{x c o s x} \prod_{i = 1}^{n} (1 - {t a n}^{2} (\frac{x}{2^{i}})) d x

Commented by ~ À ® @ 237 ~ last updated on 25/Jul/19

l e t c a l l i t F_{n} (x) . A n d P_{n} (x) = \underset{i = 1}{\prod^{n}} (1 - {t a n}^{2} (\frac{x}{2^{i}}))

w e k n o w t h a t \forall i = 1 \dots . . n 1 - {t a n}^{2} (\frac{x}{2^{i}}) = \frac{2 t a n (\frac{x}{2^{i}})}{t a n (2 . \frac{x}{2^{i}})}

s o P_{n} (x) = 2_{}^{n} \underset{i = 1}{\prod^{n}} (\frac{t a n (\frac{x}{2^{i}})}{t a n (\frac{x}{2^{i - 1}})}) = \frac{2^{n} t a n (\frac{x}{2^{n}})}{t a n x}

t h e n F_{n} (x) = \int (\frac{2^{n} t a n (\frac{x}{2^{n}})}{x s i n x}) d x = ? (i a m t r y i n g \dots . .

B u t F_{\infty} (x) = \int \frac{1}{x c o s x} P_{\infty} (x) d x i s s o l v a b l e w i t h P_{\infty} (x) = l i \underset{\infty}{m} P_{n} (x) = \frac{x}{t a n x}

t h e n

F_{\infty} (x) = \int \frac{1}{x c o s x} . \frac{x}{t a n x} d x = \int \frac{1}{s i n x} d x = \int \frac{{c o s}^{2} (\frac{x}{2}) + {s i n}^{2} (\frac{x}{2})}{2 s i n (\frac{x}{2}) c o s (\frac{x}{2})} d x

= \int \frac{\frac{1}{2} c o s (\frac{x}{2})}{s i n (\frac{x}{2})} + \frac{\frac{1}{2} s i n (\frac{x}{2})}{c o s (\frac{x}{2})} d x = l n ∣ t a n (\frac{x}{2}) ∣ + c c \in R

Commented by Prithwish sen last updated on 25/Jul/19

From

P_{n} (x) = \frac{2^{n} \tan (\frac{x}{2^{n}})}{tanx} = \frac{2^{n} \sin \frac{x}{2^{n}}}{\cos \frac{x}{2^{n}} tanx} = \frac{1}{\cos \frac{x}{2^{n}}} \underset{k = 1}{\prod^{n}} \frac{1}{\cos \frac{x}{2^{k}}}

Now if n \to \infty we know x \underset{k = 1}{\prod^{\infty} \cos} \frac{x}{2^{k}} = sinx

i . e P_{n} (x) = \frac{x}{sinx} as n \to \infty

∴ the integral becomes 2 \int \frac{dx}{\sin 2 x} = 2 \ln ∣ tanx ∣ + C

Sir please check .