1)find ∫ (√(1+t^2 )) dt 2) calculate ∫


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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
$let put I = ∫(√(1+t^2 )) dt changement t =shx give I = ∫ chx chxdx =∫ ch^2 xdx =∫ ((1+ch(2x))/2)dx = (x/2) +(1/4)sh(2x) = (x/2) + (1/2)shx chx = ((argsh(t))/2) +(1/2)t(√(1+t^2 )) =(1/2)ln(t +(√(1+t^2 )) ) +(1/2)t(√(1+t^2 )) +c 2) ∫_1 ^(√3) (√(1+t^2 )) dt =[(1/2)ln(t+(√(1+t^2 )) ) +(1/2)t(√(1+t^2 )) ]_1 ^(√3) =(1/2){ ln( 2+(√3)) +((√3)/2) .2 −ln(1+(√2)) −(√2) } =(1/2){ ln(2+(√3)) −ln(1+(√2)) +(√3) −(√2) }$
$l e t p u t I = \int \sqrt{1 + t^{2}} d t c h a n g e m e n t t = s h x g i v e$ $I = \int c h x c h x d x = \int {c h}^{2} x d x$ $= \int \frac{1 + c h (2 x)}{2} d x = \frac{x}{2} + \frac{1}{4} s h (2 x)$ $= \frac{x}{2} + \frac{1}{2} s h x c h x = \frac{a r g s h (t)}{2} + \frac{1}{2} t \sqrt{1 + t^{2}}$ $= \frac{1}{2} l n (t + \sqrt{1 + t^{2}}) + \frac{1}{2} t \sqrt{1 + t^{2}} + c$ $2) \int_{1}^{\sqrt{3}} \sqrt{1 + t^{2}} d t = {[\frac{1}{2} l n (t + \sqrt{1 + t^{2}}) + \frac{1}{2} t \sqrt{1 + t^{2}}]}_{1}^{\sqrt{3}}$ $= \frac{1}{2} {l n (2 + \sqrt{3}) + \frac{\sqrt{3}}{2} . 2 - l n (1 + \sqrt{2}) - \sqrt{2}}$ $= \frac{1}{2} {l n (2 + \sqrt{3}) - l n (1 + \sqrt{2}) + \sqrt{3} - \sqrt{2}}$
	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

	$c o r r e c t a n s w e r t h a n k s s i r M j s$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum