advanced-calculus-I-pi-2-pi-2-sin-2-tan-x-dx-pi-e-sinh-1-

Question Number 138524 by mnjuly1970 last updated on 14/Apr/21

\dots \dots . a d v a n c e d \dots \dots . \dots c a l c u l u s \dots . .

I := \int_{\frac{- π}{2}}^{\frac{π}{2}} {s i n}^{2} (t a n (x)) d x \overset{? ? ?}{=} \frac{π}{e} s i n h (1)

Answered by Dwaipayan Shikari last updated on 14/Apr/21

\int_{- \infty}^{\infty} \frac{{s i n}^{2} (t)}{t^{2} + 1} d t = τ (1)

K n o w i n g \int_{- \infty}^{\infty} \frac{c o s (α x)}{x^{2} + 1} = π e^{- α}

τ (α) = \frac{1}{2} \int_{- \infty}^{\infty} \frac{1 - c o s (2 α x)}{x^{2} + 1} d x

= \frac{1}{2} (π - π e^{- 2 α}) = \frac{π}{2} (\frac{e^{2 α} - 1}{e^{2 α}})

τ (1) = \frac{π}{2} (\frac{e^{2} - 1}{e^{2}}) = \frac{π}{2} (\frac{e - \frac{1}{e}}{e})

= \frac{π}{e} s i n h (1)

Answered by Ñï= last updated on 14/Apr/21

\int_{- π / 2}^{π / 2} \sin^{2} (\tan x) d x = \int_{- \infty}^{+ \infty} \frac{\sin^{2} u}{u^{2} + 1} d u = \int_{0}^{\infty} \frac{1 - \cos 2 u}{u^{2} + 1} d u

= \frac{π}{2} - \int_{0}^{\infty} \frac{\cos 2 u}{u^{2} + 1} d u = \frac{π}{2} - ℜ \int_{0}^{\infty} \frac{e^{i 2 u}}{u^{2} + 1} d u

= \frac{π}{2} - ℜ {π i R e s (\frac{e^{i 2 u}}{u^{2} + 1}, i)}

= \frac{π}{2} - \frac{π}{2 e^{2}}

Answered by mathmax by abdo last updated on 15/Apr/21

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{2} (tanx) dx \Rightarrow I =_{tanx = t} \int_{- \infty}^{+ \infty} \frac{\sin^{2} (t)}{1 + t^{2}} dt

= \int_{- \infty}^{+ \infty} \frac{1 - \cos (2 t)}{2 (t^{2} + 1)} dt = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dt}{t^{2} + 1} - \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{\cos (2 t)}{t^{2} + 1} dt

= \frac{π}{2} - \frac{1}{2} Re (\int_{- \infty}^{+ \infty} \frac{e^{2 it}}{t^{2} + 1} dt) let w (z) = \frac{e^{2 iz}}{z^{2} + 1} (= \frac{e^{2 iz}}{(z - i) (z + i)})

\int_{- \infty}^{+ \infty} w (z) dz = 2 i π . Res (w, i) = 2 i π . \frac{e^{- 2}}{2 i} = \frac{π}{e^{2}} \Rightarrow

I = \frac{π}{2} - \frac{π}{{2 e}^{2}} \Rightarrow I = \frac{π}{2} (1 - \frac{1}{e^{2}})