∫x^2 (√(x^2 −2))dx


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 115051 by gab last updated on 23/Sep/20

$\int x^{2} \sqrt{x^{2} - 2} d x$
Commented by Dwaipayan Shikari last updated on 23/Sep/20

$\int 2 \sqrt{2} {s i n}^{2} θ c o s θ \sqrt{2 {s i n}^{2} θ - 2} d θ x = \sqrt{2} s i n θ \Rightarrow 1 = \sqrt{2} c o s θ \frac{d θ}{d x}$ $4 i \int {s i n}^{2} θ {c o s}^{2} θ d θ$ $i \int {s i n}^{2} 2 θ d θ$ $i \int \frac{1 - c o s 4 θ}{2} d θ = \frac{θ i}{2} - \frac{1}{8} s i n 4 θ = \frac{{s i n}^{- 1} (\frac{x}{\sqrt{2}}) . i}{2} - \frac{1}{8} (4 {s i n}^{- 1} \frac{x}{\sqrt{2}}) + C$
Commented by gab last updated on 23/Sep/20

$4 i \int {s i n}^{2} θ {c o s}^{2} θ d θ ?$
Commented by gab last updated on 23/Sep/20

$s o r r y . . . i g o t i t . . . t h a n k s$
Commented by mathmax by abdo last updated on 23/Sep/20

$not correct!$
Commented by Dwaipayan Shikari last updated on 23/Sep/20

$H o w ? k i n d l y s h o w m y m i s t a k e .$
Commented by Bird last updated on 24/Sep/20

$2 {s i n}^{2} x - 2 = 2 (- {c o s}^{2} x) = - 2 {c o s}^{2} x \Rightarrow$ $\sqrt{2 {s i n}^{2} - 2} = \sqrt{2} i c o s x . . .!$
	Answered by mathmax by abdo last updated on 23/Sep/20
	$I =∫ x^2 (√(x^2 −2))dx we do the changement x =(√2)ch(t) ⇒ I =∫ 2ch^2 t (√2)sh(t)(√2)sh(t)dt =4 ∫ ch^2 (t)sh^2 (t)dt =∫ sh^2 (2t)dt =∫ ((ch(4t)−1)/2)dt =(1/8)sh(4t)−(t/2) +c =(1/8).((e^(4t) −e^(−4t) )/2)−(t/2) +c we have t =argch((x/(√2)))=ln((x/(√2))+(√((x^2 /2)−1))) ⇒I =(1/(16)){ ((x/(√2))+(√((x^2 /2)−1)))^4 −((x/2)+(√((x^2 /2)−1)))^(−4) }−(1/2)ln((x/(√2))+(√((x^2 /2)−1))) +C$
	$I = \int x^{2} \sqrt{x^{2} - 2} dx we do the changement x = \sqrt{2} ch (t) \Rightarrow$ $I = \int {2 ch}^{2} t \sqrt{2} sh (t) \sqrt{2} sh (t) dt = 4 \int {ch}^{2} (t) {sh}^{2} (t) dt$ $= \int {sh}^{2} (2 t) dt = \int \frac{ch (4 t) - 1}{2} dt = \frac{1}{8} sh (4 t) - \frac{t}{2} + c$ $= \frac{1}{8} . \frac{e^{4 t} - e^{- 4 t}}{2} - \frac{t}{2} + c we have t = argch (\frac{x}{\sqrt{2}}) = \ln (\frac{x}{\sqrt{2}} + \sqrt{\frac{x^{2}}{2} - 1})$ $\Rightarrow I = \frac{1}{16} {{(\frac{x}{\sqrt{2}} + \sqrt{\frac{x^{2}}{2} - 1})}^{4} - {(\frac{x}{2} + \sqrt{\frac{x^{2}}{2} - 1})}^{- 4}} - \frac{1}{2} \ln (\frac{x}{\sqrt{2}} + \sqrt{\frac{x^{2}}{2} - 1}) + C$
	Answered by 1549442205PVT last updated on 23/Sep/20

	$Put - {2 x}^{- 2} + 1 = t^{2} \Rightarrow 2 tdt = \frac{4}{x} dx$ $F = \int x^{2} \sqrt{x^{2} - 2} d x = \int \frac{- 2}{t^{2} - 1} \sqrt{\frac{- {2 t}^{2}}{t^{2} - 1}} \sqrt{\frac{- 2}{t^{2} - 1}} \times \frac{tdt}{2}$ $= \int \frac{- {2 t}^{2}}{{(t^{2} - 1)}^{2}} dt = \int \frac{- 2 (t^{2} - 1) - 2}{{(t^{2} - 1)}^{2}}$ $= - 2 \int \frac{dt}{t^{2} - 1} - 2 \int \frac{dt}{{(t^{2} - 1)}^{2}} = - (\int \frac{2}{t^{2} - 1} + \frac{2}{{(t^{2} - 1)}^{2}}) dt$ $= \int - (A + B) dt$ $A = \frac{2}{(t^{2} - 1)} = \frac{1}{t - 1} - \frac{1}{t + 1} . Put a = \frac{1}{t - 1}, b = \frac{1}{t + 1}$ $we get A = 2 ab = a - b$ $2 B = A^{2} = {(a - b)}^{2} = a^{2} + b^{2} - (a - b)$ $A + B = \frac{a^{2} + b^{2}}{2} + \frac{a - b}{2}$ $F = - \int (A + B) dt = - \frac{1}{2} \int (\frac{1}{{(t - 1)}^{2}} + \frac{1}{{(t + 1)}^{2}}$ $+ \frac{1}{t - 1} - \frac{1}{t + 1}) dt = \frac{1}{2 (t - 1)} + \frac{1}{2 (t + 1)}$ $+ \ln ∣ \frac{t + 1}{t - 1} ∣ + C$ $Thus, \int x^{2} \sqrt{x^{2} - 2} dx$ $= \frac{1}{2 (t - 1)} + \frac{1}{2 (t + 1)} + \ln ∣ \frac{t + 1}{t - 1} ∣ + C$ $= \frac{1}{2 \sqrt{\frac{x^{2} - 2}{x}} - 1} + \frac{1}{2 \sqrt{\frac{x^{2} - 2}{x}} + 1} + \ln ∣ \frac{\sqrt{\frac{x^{2} - 2}{x}} + 1}{2 \sqrt{\frac{x^{2} - 2}{x}} - 1} ∣ + C$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum