calculate-0-1-dx-chx-

Question Number 31515 by abdo imad last updated on 09/Mar/18

c a l c u l a t e \int_{0}^{1} \frac{d x}{c h x} .

Commented by abdo imad last updated on 10/Mar/18

I = \int_{0}^{1} \frac{d x}{\frac{e^{x} + e^{- x}}{2}} = 2 \int_{0}^{1} \frac{d x}{e^{x} + e^{- x}} t h e c h . e^{x} = t g i v e

I = 2 \int_{1}^{e} \frac{1}{t + \frac{1}{t}} \frac{d t}{t} = 2 \int_{1}^{e} \frac{d t}{t^{2} + 1} = 2 {[a r c t a n t]}_{1}^{e} = 2 a r c t a n e - \frac{π}{2} .

Answered by sma3l2996 last updated on 10/Mar/18

I = \int_{0}^{1} \frac{d x}{c o s h (x)}

t = t a n h (x / 2) \Rightarrow 2 d t = (1 - {(t a n h (x / 2))}^{2}) d x

c o s h (x) = 2 {c o s h}^{2} (x / 2) - 1 = \frac{2}{1 - {t a n h}^{2} (x / 2)} - 1

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

I = \int_{0}^{t a n h (1 / 2)} \frac{1 - t^{2}}{1 + t^{2}} \times (\frac{2 d t}{1 - t^{2}}) = 2 \int_{0}^{t a n h (1 / 2)} \frac{d t}{1 + t^{2}}

I = 2 {[{t a n}^{- 1} (t)]}_{0}^{t a n h (1 / 2)}

I = 2 {t a n}^{- 1} (t a n h (1 / 2))