calculate-0-1-1-t-2-1-2-dt-

Question Number 26055 by abdo imad last updated on 18/Dec/17

c a l c u l a t e \int_{0}^{1} {(1 + t^{2})}^{1 / 2} d t

Answered by Joel578 last updated on 19/Dec/17

I = \int_{0}^{1} \sqrt{t^{2} + 1} d t

Let t = \tan x \to d t = \sec^{2} x d x

t = 0 \to x = 0

t = 1 \to x = \frac{π}{4}

I = \int_{0}^{π / 4} \sqrt{\tan^{2} x + 1} . \sec^{2} x d x

= \int_{0}^{π / 4} \sec^{3} x d x

Using I B P give :

= {[\frac{1}{2} (\sec x \tan x) - \frac{1}{2} \ln ∣ \sec x + \tan x ∣]}_{0}^{π / 4}

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

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

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