let-0-pi-calculate-0-pi-2-dx-2-cos-chx-

Question Number 30775 by abdo imad last updated on 25/Feb/18

l e t α \in] 0, π [c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d x}{2 (c o s α + c h x)} .

Commented by prof Abdo imad last updated on 25/Feb/18

w e h a v e I = \int_{0}^{\frac{π}{2}} \frac{d x}{2 (c o s α + \frac{e^{x} + e^{- x}}{2})}

= \int_{0}^{\frac{π}{2}} \frac{d x}{2 c o s α + e^{x} + e^{- x}} t h e c h . e^{x} = t

I = \int_{1}^{e^{\frac{π}{2}}} \frac{1}{2 c o s α + t + \frac{1}{t}} \frac{d t}{t}

= \int_{1}^{e^{\frac{π}{2}}} \frac{d t}{2 t c o s α + t^{2} + 1} = \int_{1}^{e^{\frac{π}{2}}} \frac{d t}{t^{2} + 2 t c o s α + 1}

= \int_{0}^{e^{\frac{π}{2}}} \frac{d t}{{(t + c o s α)}^{2} + {s i n}^{2} α} l e t u s e t h e c h .

t + c o s α = u s i n α \Rightarrow u = \frac{1}{s i n α} (t + c o s α)

I = \int_{c o t a n (α)}^{\frac{e^{\frac{π}{2}} + c o s α}{s i n α}} \frac{s i n α d u}{{s i n}^{2} α (1 + u^{2})}

= \frac{1}{s i n α} {[a r c t a n u]}_{c o t a n α}^{\frac{e^{\frac{π}{2}} + c o s α}{s i n α}}

= \frac{1}{s i n α} (a r c t a n (\frac{e^{\frac{π}{2}} + c o s α}{s i n α}) - a r c t a n (c o t a n α)) .

Answered by sma3l2996 last updated on 25/Feb/18

A = \int_{0}^{π / 2} \frac{d x}{2 (c o s α + c h x)}

l e t t = t h (x / 2) \Rightarrow d t = \frac{1}{2} (1 - {t h}^{2} (x / 2)) d x

d x = \frac{2}{1 - t^{2}} d t

c h (x) = 2 {c h}^{2} (x / 2) - 1 = \frac{2}{1 - {t h}^{2} (x / 2)} - 1 = \frac{2}{1 - t^{2}} - 1 = \frac{1 + t^{2}}{1 - t^{2}}

A = \int_{0}^{t h (π / 4)} \frac{1}{2 (c o s α + \frac{1 + t^{2}}{1 - t^{2}})} \times (\frac{2 d t}{1 - t^{2}})

= \int_{0}^{t h (π / 4)} \frac{d t}{(1 - t^{2}) c o s α + 1 + t^{2}} = \int_{0}^{t h (π / 4)} \frac{d t}{t^{2} (1 - c o s α) + 1 + c o s α}

A = \int_{0}^{t h (π / 4)} \frac{d t}{(1 + c o s α) ((\frac{1 - c o s α}{1 + c o s α}) t^{2} + 1)}

{l e t}^{'} s p u t u = {(\frac{1 - c o s α}{1 + c o s α})}^{1 / 2} t \Rightarrow d t = (\frac{1 + c o s α}{1 - c o s α}) d u

A = \frac{1}{{(1 - c o s α)}^{1 / 2} {(1 + c o s α)}^{1 / 2}} \int_{0}^{a} \frac{d u}{u^{2} + 1}

A = \frac{1}{{(1 - {c o s}^{2} α)}^{1 / 2}} \times {[a r c t a n (\frac{1 - c o s α}{s i n α} t)]}_{0}^{t h (π / 4)}

A = \frac{1}{s i n α} a r c t a n (\frac{1 - c o s α}{s i n α} t h (π / 4))