∫ (x^2 /( (√((4−x^2 )^5 )))) dx


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Question Number 119262 by bemath last updated on 23/Oct/20

$\int \frac{x^{2}}{\sqrt{{(4 - x^{2})}^{5}}} d x$
	Answered by benjo_mathlover last updated on 23/Oct/20

	$s e t x = 2 \sin r \Rightarrow d x = 2 \cos r d r$ $\int \frac{4 \sin^{2} r . 2 \cos r d r}{{(\sqrt{4 - 4 \sin^{2} r})}^{5}} =$ $\int \frac{\sin^{2} r . \cos r d r}{4 . \cos^{5} r} =$ $\frac{1}{4} \int \frac{\tan^{2} r}{\cos^{2} r} d r = \frac{1}{4} \int \tan^{2} r \sec^{2} r d r$ $= \frac{1}{12} \tan^{3} r + c$ $= \frac{x^{3}}{12 \sqrt{{(4 - x^{2})}^{3}}} + c$
	Answered by mathmax by abdo last updated on 23/Oct/20
	$I=∫ (x^2 /((4−x^2 )^(5/2) ))dx we do the changement x=2sint ⇒ I =∫ ((4sin^2 t)/((4cos^2 t)^(5/2) ))(2cost)dt =(8/4^(5/2) )∫ ((sin^2 t)/(cos^4 t))dt we have ∫ ((sin^2 t)/(cos^4 t)) dt =∫ ((1−cos^2 t)/(cos^4 t))dt =∫ (dt/(cos^4 t))−∫ (dt/(cos^2 t)) ∫ (dt/(cos^2 t)) =tant +c_1 =tan(arcsin((x/2)))+c_1 ∫ (dt/(cos^4 t)) =∫ (dt/((((1+cos(2t))/2))^2 )) =∫ ((4dt)/(1+2cos(2t)+cos^2 (2t))) =∫ ((4dt)/(1+2cos(2t)+((1+cos(4t))/2))) =∫ ((8dt)/(2+4cos(2t)+1+cos(4t))) =∫ ((8dt)/(3+4cos(2t)+2cos^2 (2t)−1)) =_(2t=u) ∫ ((4du)/(2+4cosu+2cos^2 u)) =2∫ (du/(1+2cosu +cos^2 u)) =2 ∫ (du/((1+cosu)^2 )) =_(tan((u/2))=y) =2∫ ((2dy)/((1+y^2 )(1+((1−y^2 )/(1+y^2 )))^2 )) =4∫ (dy/((1+y^2 )×(4/((1+y^2 )^2 )))) =∫ (1+y^2 )dy =y+(y^3 /3) =tan((u/2))+(1/3)tan^3 ((u/2))+c_2 =tan(t)+(1/3)tan^3 (t)+c_2 =tan(arcsn((x/2)))+(1/3)(arcsin((x/2)))^3 +c_2 ⇒I =(2^3 /2^5 ){tan(arcsin((x/2)))+(1/3)(tanarcsin((x/2)))^3 −tan(arcsin((x/2)))+C =(1/(12))(tan(arcsin((x/2)))^3 ) +C$
	$I = \int \frac{x^{2}}{{(4 - x^{2})}^{\frac{5}{2}}} dx we do the changement x = 2 sint \Rightarrow$ $I = \int \frac{{4 \sin}^{2} t}{{({4 \cos}^{2} t)}^{\frac{5}{2}}} (2 cost) dt = \frac{8}{4^{\frac{5}{2}}} \int \frac{\sin^{2} t}{\cos^{4} t} dt we have$ $\int \frac{\sin^{2} t}{\cos^{4} t} dt = \int \frac{1 - \cos^{2} t}{\cos^{4} t} dt = \int \frac{dt}{\cos^{4} t} - \int \frac{dt}{\cos^{2} t}$ $\int \frac{dt}{\cos^{2} t} = tant + c_{1} = \tan (\arcsin (\frac{x}{2})) + c_{1}$ $\int \frac{dt}{\cos^{4} t} = \int \frac{dt}{{(\frac{1 + \cos (2 t)}{2})}^{2}} = \int \frac{4 dt}{1 + 2 \cos (2 t) + \cos^{2} (2 t)}$ $= \int \frac{4 dt}{1 + 2 \cos (2 t) + \frac{1 + \cos (4 t)}{2}} = \int \frac{8 dt}{2 + 4 \cos (2 t) + 1 + \cos (4 t)}$ $= \int \frac{8 dt}{3 + 4 \cos (2 t) + {2 \cos}^{2} (2 t) - 1} =_{2 t = u} \int \frac{4 du}{2 + 4 cosu + {2 \cos}^{2} u}$ $= 2 \int \frac{du}{1 + 2 cosu + \cos^{2} u} = 2 \int \frac{du}{{(1 + cosu)}^{2}} =_{\tan (\frac{u}{2}) = y}$ $= 2 \int \frac{2 dy}{(1 + y^{2}) {(1 + \frac{1 - y^{2}}{1 + y^{2}})}^{2}} = 4 \int \frac{dy}{(1 + y^{2}) \times \frac{4}{{(1 + y^{2})}^{2}}}$ $= \int (1 + y^{2}) dy = y + \frac{y^{3}}{3} = \tan (\frac{u}{2}) + \frac{1}{3} \tan^{3} (\frac{u}{2}) + c_{2}$ $= \tan (t) + \frac{1}{3} \tan^{3} (t) + c_{2} = \tan (arcsn (\frac{x}{2})) + \frac{1}{3} {(\arcsin (\frac{x}{2}))}^{3} + c_{2}$ $\Rightarrow I = \frac{2^{3}}{2^{5}} {\tan (\arcsin (\frac{x}{2})) + \frac{1}{3} {(tanarcsin (\frac{x}{2}))}^{3} - \tan (\arcsin (\frac{x}{2})) + C$ $= \frac{1}{12} (\tan {(\arcsin (\frac{x}{2}))}^{3}) + C$
	Answered by MJS_new last updated on 23/Oct/20

	$without trigonometric substitution$ $\int \frac{x^{2}}{{(4 - x^{2})}^{5 / 2}} d x =$ $[t = \frac{x}{{(4 - x^{2})}^{1 / 2}} \Leftrightarrow x = \frac{2 t}{\sqrt{t^{2} + 1}} \to d x = \frac{{(4 - x^{2})}^{3 / 2}}{4} d t]$ $= \frac{1}{4} \int t^{2} d t = \frac{1}{12} t^{3} = \frac{x^{3}}{12 {(4 - x^{3})}^{3 / 2}} + C$
	Commented by benjo_mathlover last updated on 23/Oct/20

	$a m a z i n g . . . s i r$
	Commented by MJS_new last updated on 23/Oct/20

	$just decades of experience$
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