calculate-1-2-dx-ch-2-x-sh-2-x-

Question Number 115366 by Bird last updated on 25/Sep/20

c a l c u l a t e \int_{- 1}^{2} \frac{d x}{{c h}^{2} x + {s h}^{2} x}

Answered by MJS_new last updated on 25/Sep/20

\int \frac{d x}{\cosh^{2} x + \sinh^{2} x} =

[t = e^{2 x} \to d x = \frac{d t}{{2 e}^{2 x}}]

= \int \frac{d t}{t^{2} + 1} = \arctan t =

= \arctan e^{2 x} + C

\underset{- 1}{\int^{2}} \frac{d x}{\cosh^{2} x + \sinh^{2} x} = \arctan e^{4} - \arctan e^{- 2}

Commented by mathmax by abdo last updated on 25/Sep/20

thank you sir mjs

Answered by mathmax by abdo last updated on 25/Sep/20

let I = \int_{- 1}^{2} \frac{dx}{{ch}^{2} x + {sh}^{2} x} \Rightarrow I = \int_{- 1}^{2} \frac{dx}{\frac{1 + ch (2 x)}{2} + \frac{ch (2 x) - 1}{2}}

= \int_{- 1}^{2} \frac{dx}{ch (2 x)} = 2 \int_{- 1}^{2} \frac{dx}{e^{2 x} + e^{- 2 x}} =_{e^{2 x} = t} 2 \int_{e^{- 2}}^{e^{- 4}} \frac{1}{t + t^{- 1}} \times \frac{dt}{2 t}

= \int_{e^{- 2}}^{e^{- 4}} \frac{dt}{t^{2} + 1} = \arctan (\frac{1}{e^{4}}) - \arctan (\frac{1}{e^{2}})

= \frac{π}{2} - \arctan (e^{4}) - (\frac{π}{2} - \arctan (e^{2})) = \arctan (e^{2}) - \arctan (e^{4})