find-1-1-dt-t-1-t-2-

Question Number 28885 by abdo imad last updated on 31/Jan/18

f i n d \int_{- 1}^{1} \frac{d t}{t + \sqrt{1 + t^{2}}} .

Commented by abdo imad last updated on 02/Feb/18

l e t p u t I = \int_{- 1}^{1} \frac{d t}{t + \sqrt{1 + t^{2}}}

I = \int_{- 1}^{1} \frac{\sqrt{1 + t^{2}} - t}{1 + t^{2} - t^{2}} d t = \int_{- 1}^{1} \sqrt{1 + t^{2} d t} - \int_{- 1}^{1} t d t

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

\int_{0}^{1} \sqrt{1 + t^{2}} d t = \int_{0}^{\frac{π}{4}} c o s x (1 + {t a n}^{2} x) d x = \int_{0}^{\frac{π}{4}} \frac{d x}{c o s x} a n d t h e c h .

t a n (\frac{x}{2}) = u g i v e

\int_{0}^{\frac{π}{4}} \frac{d x}{c o s x} = \int_{0}^{\sqrt{2} - 1} \frac{1}{\frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = \int_{0}^{\sqrt{2} - 1} \frac{2 d u}{1 - u^{2}}

= \int_{0}^{\sqrt{2} - 1} (\frac{1}{1 + u} + \frac{1}{1 - u}) d u = {[l n ∣ \frac{1 + u}{1 - u} ∣]}_{0}^{\sqrt{2} - 1}

= l n (\frac{\sqrt{2}}{2 - \sqrt{2}}) = l n (\frac{\sqrt{2}}{\sqrt{2} (\sqrt{2} - 1)}) = - l n (\sqrt{2} - 1) .

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