1-find-1-t-2-dt-2-calculate-1-3-1-t-2-dt-

Question Number 35044 by math khazana by abdo last updated on 14/May/18

1) f i n d \int \sqrt{1 + t^{2}} d t

2) c a l c u l a t e \int_{1}^{\sqrt{3}} \sqrt{1 + t^{2}} d t

Commented by math khazana by abdo last updated on 15/May/18

l e t p u t I = \int \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 c h x c h x d x = \int {c h}^{2} x d x

= \int \frac{1 + c h (2 x)}{2} d x = \frac{x}{2} + \frac{1}{4} s h (2 x)

= \frac{x}{2} + \frac{1}{2} s h x c h x = \frac{a r g s h (t)}{2} + \frac{1}{2} t \sqrt{1 + t^{2}}

= \frac{1}{2} l n (t + \sqrt{1 + t^{2}}) + \frac{1}{2} t \sqrt{1 + t^{2}} + c

2) \int_{1}^{\sqrt{3}} \sqrt{1 + t^{2}} d t = {[\frac{1}{2} l n (t + \sqrt{1 + t^{2}}) + \frac{1}{2} t \sqrt{1 + t^{2}}]}_{1}^{\sqrt{3}}

= \frac{1}{2} {l n (2 + \sqrt{3}) + \frac{\sqrt{3}}{2} . 2 - l n (1 + \sqrt{2}) - \sqrt{2}}

= \frac{1}{2} {l n (2 + \sqrt{3}) - l n (1 + \sqrt{2}) + \sqrt{3} - \sqrt{2}}

Answered by MJS last updated on 14/May/18

\int \sqrt{1 + t^{2}} d t =

[t = \tan (u) \to d t = \sec^{2} (u) d u]

= \int \sec^{2} (u) \sqrt{1 + \tan^{2} (u)} d u =

= \int \sec^{3} (u) d u =

[\int \sec^{n} (u) d u = \frac{\sec^{n - 2} (u) \tan (u)}{n - 1} + \frac{n - 2}{n - 1} \int \sec^{n - 2} (u) d u]

= \frac{\sec (u) \tan (u)}{2} + \frac{1}{2} \int \sec (u) d u =

= \frac{\sec (u) \tan (u)}{2} + \frac{1}{2} \int \sec (u) \frac{\tan (u) + \sec (u)}{\tan (u) + \sec (u)} d u =

= \frac{\sec (u) \tan (u)}{2} + \frac{1}{2} \int \frac{\tan (u) \sec (u) + \sec^{2} (u)}{\tan (u) + \sec (u)} d u =

[v = \tan (u) + \sec (u) \to d u = \frac{d v}{\tan (u) \sec (u) + \sec^{2} (u)}]

= \frac{\sec (u) \tan (u)}{2} + \frac{1}{2} \int \frac{1}{v} d v =

= \frac{\sec (u) \tan (u)}{2} + \frac{\ln (v)}{2} + C =

= \frac{\sec (u) \tan (u)}{2} + \frac{\ln ∣ \tan (u) + \sec (u) ∣}{2} + C =

= \frac{\sec (\arctan (t)) \tan (arcran (t))}{2} + \frac{\ln ∣ \tan (\arctan (t)) + \sec (\arctan (t)) ∣}{2} + C =

= \frac{t \sqrt{1 + t^{2}}}{2} + \frac{\ln ∣ t + \sqrt{1 + t^{2}} ∣}{2} + C

\underset{1}{\int^{\sqrt{3}}} \sqrt{1 + t^{2}} d t = {[\frac{t \sqrt{1 + t^{2}}}{2} + \frac{\ln ∣ t + \sqrt{1 + t^{2}} ∣}{2}]}_{1}^{\sqrt{3}} =

= \sqrt{3} + \frac{\ln (2 + \sqrt{3})}{2} - (\frac{\sqrt{2}}{2} + \frac{\ln (1 + \sqrt{2})}{2}) =

= \sqrt{3} - \frac{\sqrt{2}}{2} + \frac{1}{2} \ln (\frac{2 + \sqrt{3}}{1 + \sqrt{2}}) = \sqrt{3} - \frac{\sqrt{2}}{2} + \frac{1}{2} \ln (\sqrt{6} - \sqrt{3} + 2 \sqrt{2} - 2) \approx

\approx 1.242736

Commented by math khazana by abdo last updated on 15/May/18

c o r r e c t a n s w e r t h a n k s s i r M j s