nice-calculus-prove-1-0-pi-2-tan-1-tan-x-tan-x-dx-pi-2-log-2-2-2-pi-pi-e-sin-x-cos-x-cos-sin-x-e-x-e-sin-x-dx-pi-

Question Number 136295 by mnjuly1970 last updated on 20/Mar/21

\dots . n i c e c a l c u l u s \dots

p r o v e ::

1 :: ϕ = \int_{0}^{\frac{π}{2}} \frac{{t a n}^{- 1} (\sqrt{t a n (x)})}{t a n (x)} d x = \frac{π}{2} l o g (2 + \sqrt{2})

2 :: Ω = \int_{- π}^{π} \frac{e^{(s i n (x) + c o s (x))} c o s (s i n (x))}{e^{x} + e^{s i n (x)}} d x = π

Answered by mindispower last updated on 20/Mar/21

Ω = R e \int_{- π}^{π} \frac{e^{c o s (x) + s i n (x) (+ i s i n (x)}}{e^{x} + s i n (x)} d x = R e (W)

W = \int_{- π}^{π} \frac{e^{c o s (x) + i s i n (x)}}{e^{x - s i n (x)} + 1} d x, z = c o s (x) + i s i n (x), \bar{z}, c o n j (z)

b y s e m e t r y x \to - x

2 w = \int_{- π}^{π} \frac{e^{\bar{z}}}{e^{- x + s i n (x)} + 1} + \frac{e^{z}}{e^{x - s i n (x)} + 1} d x

= \int_{- π}^{π} \frac{e^{\bar{z}} (e^{x - s i n (x)}) + e^{z}}{e^{x - s i n (x)} + 1} d x

= \int_{- π}^{π} e^{\bar{z}} + \int_{- π}^{π} \frac{e^{z} - e^{\bar{z}}}{e^{x - s i n (x)} + 1} d x

2 w = A + B

B \in {i a, a \in R}

Ω = R e (w) = \frac{1}{2} R e \int_{- π}^{π} e^{\bar{z}} d x, x \to - x, s i n (- x) = - s i n (x) \Rightarrow

= \frac{1}{2} R e \int_{- π}^{π} e^{z} d x = \frac{1}{2} R e \sum_{n ⩾ 0} \int_{- π}^{π} \frac{z^{n}}{n!} d x

z = e^{i x}

Missing \left or extra \right

x \to e^{i n x} i s 2 π p e r i o d i c \Rightarrow D = 0

Ω = \frac{1}{2} . 2 π + 0 = π

\int_{- π}^{π} \frac{e^{c o s (x) + s i n (x)} c o s (s i n (x)}{e^{x} + e^{s i n (x)}} d x = π

Commented by mnjuly1970 last updated on 20/Mar/21

v e r y n i c e s o l u t i o n

t h a n k s a l o t \dots

Commented by mindispower last updated on 20/Mar/21

t h a n x a l w a y s p l e a s u r

Answered by Ar Brandon last updated on 20/Mar/21

ϕ = \int_{0}^{\frac{π}{2}} \frac{\tan^{- 1} (\sqrt{tanx})}{tanx} dx \underset{u = tanx}{=} \int_{0}^{\infty} \frac{\tan^{- 1} (\sqrt{u})}{u (1 + u^{2})} du

= \int_{0}^{\infty} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} u^{n}}{(n + 1) (1 + u^{2})} du = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + 1} \int_{0}^{\infty} \frac{u^{n}}{(1 + u^{2})} du

\underset{t = u^{2}}{=} \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + 1} \int_{0}^{\infty} \frac{t^{\frac{n}{2} - \frac{1}{2}}}{(1 + t)} dt = \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + 1} β (\frac{n}{2} + \frac{1}{2}, \frac{1}{2} - \frac{n}{2})

= \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + 1} \cdot \frac{π}{\cos (\frac{π n}{2})}

Commented by mindispower last updated on 20/Mar/21

y o u c a n t [s w i t c h Σ a n d \int