calculate-1-dt-t-1-t-2-

Question Number 36182 by prof Abdo imad last updated on 30/May/18

c a l c u l a t e \int_{1}^{+ \infty} \frac{d t}{t \sqrt{1 + t^{2}}}

Commented by maxmathsup by imad last updated on 15/Aug/18

l e t I = \int_{1}^{+ \infty} \frac{d t}{t \sqrt{1 + t^{2}}} d t c h a n g e m e n t t = s h (x) g i v e

I = \int_{a r g s h (1)}^{+ \infty} \frac{1}{s h (x) c h (x)} c h (x) d x = \int_{l n (1 + \sqrt{2})}^{+ \infty} \frac{d x}{s h (x)}

= \int_{l n (1 + \sqrt{2})}^{+ \infty} \frac{2 d x}{e^{x} - e^{- x}} = \int_{l n (1 + \sqrt{2})}^{+ \infty} \frac{2 e^{x}}{e^{2 x} - 1} d x =_{e^{x} = u} \int_{1 + \sqrt{2}}^{+ \infty} \frac{2 u}{u^{2} - 1} \frac{d u}{u}

= \int_{1 + \sqrt{2}}^{+ \infty} \frac{2 d u}{u^{2} - 1} = \int_{1 + \sqrt{2}}^{+ \infty} (\frac{1}{u - 1} - \frac{1}{u + 1}) d u

= {[l n ∣ \frac{u - 1}{u + 1} ∣]}_{1 + \sqrt{2}}^{+ \infty} = - l n ∣ \frac{\sqrt{2}}{2 + \sqrt{2}} ∣ = - l n (\frac{\sqrt{2}}{\sqrt{2 (1 + \sqrt{2})}}) = l n (1 + \sqrt{2}) \Rightarrow

I = l n (1 + \sqrt{2}) .

Answered by sma3l2996 last updated on 31/May/18

L = \int_{1}^{+ \infty} \frac{d t}{t \sqrt{1 + t^{2}}}

l e t t = s i n h x \Rightarrow d t = c o s h x d x

L = \int_{l n (\sqrt{2} + 1)}^{+ \infty} \frac{c o s h x}{s i n h x \sqrt{1 + {s i n h}^{2} x}} d x

= \int_{l n (\sqrt{2} + 1)}^{+ \infty} \frac{d x}{s i n h (x)} = 2 \int_{l n (\sqrt{2} + 1)}^{+ \infty} \frac{e^{x}}{e^{2 x} - 1} d x

u = e^{x} \Rightarrow d u = e^{x} d x

L = 2 \int_{\sqrt{2} + 1}^{+ \infty} \frac{d u}{u^{2} - 1} = 2 \int_{\sqrt{2} + 1}^{+ \infty} \frac{d u}{(u + 1) (u - 1)}

= \int_{\sqrt{2} + 1}^{+ \infty} (\frac{1}{u - 1} - \frac{1}{u + 1}) d u

= {[l n ∣ \frac{u - 1}{u + 1} ∣]}_{\sqrt{2} + 1}^{+ \infty} = l n ∣ \frac{\sqrt{2} + 2}{\sqrt{2}} ∣=

L = l n (\sqrt{2} + 1)

Answered by ajfour last updated on 31/May/18

l e t t = \tan θ; t h e n

I = \int_{π / 4}^{π / 2} \frac{\sec^{2} θ d θ}{(\tan θ) \sec θ}

I = \int_{π / 4}^{π / 2} cosec θ d θ

= \ln ∣ \frac{cosec (π / 2) - \cot (π / 2)}{cosec (π / 4) - \cot (π / 4)} ∣

= \ln (\frac{1}{\sqrt{2} - 1}) = \ln (1 + \sqrt{2}) .

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