∫(√((2+x)/(2−x)))dx


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 136907 by leena12345 last updated on 27/Mar/21

$\int \sqrt{\frac{2 + x}{2 - x}} d x$
	Answered by Mathspace last updated on 27/Mar/21
	$let I=∫(√((2+x)/(2−x)))dx changement (√((2+x)/(2−x)))=t give ((2+x)/(2−x))=t^(2 ) ⇒ 2+x=2t^2 −t^2 x ⇒(1+t^2 )x=2t^2 −2 ⇒x=((2t^2 −2)/(t^(2 ) +1)) ⇒(dx/dt)=((4t(t^2 +1)−2t(2t^2 −2))/((t^2 +1)^2 )) =((8t)/((t^2 +1)^2 )) ⇒ I=∫ t×((8t)/((t^2 +1)^2 ))dt =4 ∫ ((2t)/((t^2 +1)^2 ))×tdt by parts f^(′ ) =((2t)/((t^2 +1)^2 )) and g=t ⇒ I=4{−(t/((t^2 +1)))+∫ (1/((t^2 +1)))dt} =((−4t)/(t^2 +1)) +4arctant +C =−4((√((2+x)/(2−x)))/(((2+x)/(2−x))+1)) +4arctan(√((2+x)/(2−x))) +C =−4(√((2+x)/(2−x))).((2−x)/4)+4arctan(√((2+x)/(2−x)))+C =−(√(4−x^2 ))+4arctan((√((2+x)/(2−x)))) +C$
	$l e t I = \int \sqrt{\frac{2 + x}{2 - x}} d x c h a n g e m e n t$ $\sqrt{\frac{2 + x}{2 - x}} = t g i v e \frac{2 + x}{2 - x} = t^{2} \Rightarrow$ $2 + x = 2 t^{2} - t^{2} x \Rightarrow (1 + t^{2}) x = 2 t^{2} - 2$ $\Rightarrow x = \frac{2 t^{2} - 2}{t^{2} + 1} \Rightarrow \frac{d x}{d t} = \frac{4 t (t^{2} + 1) - 2 t (2 t^{2} - 2)}{{(t^{2} + 1)}^{2}}$ $= \frac{8 t}{{(t^{2} + 1)}^{2}} \Rightarrow$ $I = \int t \times \frac{8 t}{{(t^{2} + 1)}^{2}} d t$ $= 4 \int \frac{2 t}{{(t^{2} + 1)}^{2}} \times t d t$ $b y p a r t s f^{^{'}} = \frac{2 t}{{(t^{2} + 1)}^{2}} a n d g = t \Rightarrow$ $I = 4 {- \frac{t}{(t^{2} + 1)} + \int \frac{1}{(t^{2} + 1)} d t}$ $= \frac{- 4 t}{t^{2} + 1} + 4 a r c t a n t + C$ $= - 4 \frac{\sqrt{\frac{2 + x}{2 - x}}}{\frac{2 + x}{2 - x} + 1} + 4 a r c t a n \sqrt{\frac{2 + x}{2 - x}} + C$ $= - 4 \sqrt{\frac{2 + x}{2 - x}} . \frac{2 - x}{4} + 4 a r c t a n \sqrt{\frac{2 + x}{2 - x}} + C$ $= - \sqrt{4 - x^{2}} + 4 a r c t a n (\sqrt{\frac{2 + x}{2 - x}}) + C$
	Answered by Dwaipayan Shikari last updated on 27/Mar/21

	$\int \frac{2 + x}{\sqrt{4 - x^{2}}} d x$ $= 2 \int \frac{1}{\sqrt{4 - x^{2}}} - \frac{1}{2} \int \frac{- 2 x}{\sqrt{4 - x^{2}}} d x$ $= 2 {s i n}^{- 1} \frac{x}{2} - \sqrt{4 - x^{2}} + C$
	Answered by Mathspace last updated on 27/Mar/21

	$a n o t h e r w a y f o r t h i s k i n d o f \int$ $w e d o c h a n g e m e n t x = 2 c o s θ \Rightarrow$ $\int \sqrt{\frac{2 + x}{2 - x}} d x = \int \sqrt{\frac{2 + 2 c o s θ}{2 - 2 c o s θ}} (- 2 s i n θ) d θ$ $= - 2 \int \sqrt{\frac{1 + c o s θ}{1 - c o s θ}} s i n θ$ $= - 2 \int \frac{c o s (\frac{θ}{2})}{s i n (\frac{θ}{2})} (2 c o s (\frac{θ}{2}) s i n (\frac{θ}{2}) d θ$ $= - 4 \int {c o s}^{2} (\frac{θ}{2}) d θ$ $= - 2 \int (1 + c o s θ) d θ$ $= - 2 θ - 2 s i n θ + C$ $= - 2 a r c o s (\frac{x}{2}) - 2 \sqrt{1 - \frac{x^{2}}{4}} + C$
	Answered by liberty last updated on 27/Mar/21

	$I = \int \sqrt{\frac{2 + x}{2 - x}} dx$ $let x = 2 \cos 2 α \Rightarrow dx = - 4 \sin 2 α d α$ $I = - 4 \int \sin 2 α \sqrt{\frac{2 + 2 \cos 2 α}{2 - 2 \cos 2 α}} d α$ $I = - 4 \int \sin 2 α \sqrt{\frac{2 (2 \sin^{2} α)}{2 (2 \cos^{2} α)}} d α$ $I = - 4 \int \sin 2 α (\frac{\sin α}{\cos α}) d α$ $I = - 8 \int \sin^{2} α d α$ $I = \int (- 4 + 4 \cos 2 α) d α$ $I = - 4 α + 2 \sin 2 α + c$ $I = - 4 (\frac{\arccos (\frac{x}{2})}{2}) + 2 \sqrt{1 - \frac{x^{2}}{4}} + c$ $I = - 2 \arccos (\frac{x}{2}) + \sqrt{4 - x^{2}} + c$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum