Evaluate-dx-x-4-9-x-2-

Question Number 179405 by Acem last updated on 29/Oct/22

E v a l u a t e \int \frac{d x}{x^{4} \sqrt{9 - x^{2}}}

Answered by greougoury555 last updated on 29/Oct/22

M = \int \frac{d x}{x^{3} x^{2} \sqrt{\frac{9}{x^{2}} - 1}} d x

M = \int \frac{x^{- 3}}{x^{2} \sqrt{9 x^{- 2} - 1}} d x

l e t j^{2} = 9 x^{- 2} - 1 \Rightarrow x^{- 2} = \frac{1}{9} (j^{2} + 1)

2 j d j = - 18 x^{- 3} d x

M = \int \frac{(- \frac{1}{9} j) d j}{\frac{1}{9} (j^{2} + 1) . j} = - \int \frac{d j}{j^{2} + 1}

M = - \arctan (\sqrt{\frac{9}{x^{2}} - 1}) + c

M = - \arctan (\frac{x}{\sqrt{9 - x^{2}}}) + c

Commented by Acem last updated on 29/Oct/22

M = \int \frac{x^{- 3}}{x^{2} \sqrt{9 x^{- 2} - 1}} d x

M = \int \frac{(- \frac{1}{9} j) d j}{\frac{1}{9} (j^{2} + 1) . j} = - \int \frac{d j}{j^{2} + 1}

x^{2} \neq \frac{1}{9} (j^{2} + 1)

Commented by greougoury555 last updated on 29/Oct/22

h m m m \dots x^{- 2} \bar{} n o t x^{2}

Commented by Acem last updated on 29/Oct/22

s u r e x^{- 2}, a n d s o m e t h i n g w i l l c h a n g e

Commented by Acem last updated on 29/Oct/22

B y t h e w a y, y o u r m e t h o d i s v e r y g o o d!

Commented by Acem last updated on 29/Oct/22

W h e n y o u s u p p o s e d j^{2} = 9 x^{- 2} - 1 y o u g o t

j d j = - 9 x^{- 3} d x, x^{- 2} = \frac{1}{9} (j^{2} + 1)

y o u h a d t o w r i t e M = \int \frac{x^{- 3} x^{- 2}}{\sqrt{9 x^{- 2} - 1}} d x t h e n

M = \frac{- 1}{81} \int (j^{2} + 1) d j = \frac{- 1}{81} (\frac{1}{3} j^{3} + j) + c

M = \frac{- 1}{81} (\frac{\sqrt{{(9 - x^{2})}^{3}}}{3 x^{3}} + \frac{\sqrt{9 - x^{2}}}{x}) + c

Answered by ARUNG_Brandon_MBU last updated on 29/Oct/22

I = \int \frac{d x}{x^{4} \sqrt{9 - x^{2}}} = \int \frac{d x}{x^{5} \sqrt{\frac{9}{x^{2}} - 1}}, t = \frac{1}{x^{2}} \Rightarrow d t = - \frac{2}{x^{3}} d x

= - \frac{1}{2} \int \frac{t}{\sqrt{9 t - 1}} d t = - \frac{1}{18} \int \frac{9 t - 1 + 1}{\sqrt{9 t - 1}} d t

= - \frac{1}{18} \int (\sqrt{9 t - 1} + \frac{1}{\sqrt{9 t - 1}}) d t

= - \frac{1}{162} (\frac{2}{3} {(9 t - 1)}^{\frac{3}{2}} + 2 \sqrt{9 t - 1}) + C

= - \frac{1}{81} (\frac{\sqrt{{(9 - x^{2})}^{3}}}{3 x^{3}} + \frac{\sqrt{9 - x^{2}}}{x}) + C

Commented by Acem last updated on 29/Oct/22

{I t}^{'} s v e r y g o o d s u b s t i t u t i o n!! T h a n k s

Answered by ARUNG_Brandon_MBU last updated on 29/Oct/22

I = \int \frac{d x}{x^{4} \sqrt{9 - x^{2}}}, x = 3 \sin θ

= \frac{1}{81} \int \frac{\cos θ}{\sin^{4} θ ∣ \cos θ ∣} d θ = \frac{sgn (\cos θ)}{81} \int \frac{d θ}{\sin^{4} θ}

= \frac{sgn (\cos θ)}{81} \int {cosec}^{2} θ (\cot^{2} θ + 1) d θ

= - \frac{sgn (\cos θ)}{81} (\frac{\cot^{3} θ}{3} + \cot θ) + C

Commented by Acem last updated on 29/Oct/22

T h i s i s m y m e t h o d . B y t h e w a y a s l o n g a s

t h e i n t e g r a t i o n i s u n d e f i n e d, w e c a n d e e m t h a t

∣ \cos θ ∣= \cos θ b u t i f i t w a s l i m i t w e h a v e t o c o m p t

t h e l i m i t s a n d s e e i t d o m a i n

i f \frac{- π}{2} ⩽ θ ⩽ \frac{π}{2} t h e n ∣ \cos θ ∣= \cos θ, o t h e r w i s e - \cos θ

T h a n k s f o r y o u

Commented by ARUNG_Brandon_MBU last updated on 29/Oct/22

OK thanks

Answered by Acem last updated on 29/Oct/22

x = 3 \sin θ \Rightarrow d x = 3 \cos θ d θ

\sqrt{9 - x^{2}} = 3 \sqrt{\cos^{2} θ} = 3 ∣ \cos θ ∣= 3 \cos θ^{I n t e g . u n d e f i n e d}

I = \int \frac{3 \cos θ}{81 \sin^{4} θ 3 \cos θ} d θ = \frac{1}{81} \int {cosec}^{2} θ (\cot^{2} θ + 1) d θ

= \frac{- 1}{81} (\frac{1}{3} \cot^{3} θ + \cot θ) + c; {\begin{cases} {(\cot θ)}^{^{'}} = - {cosec}^{2} θ \\ \int {cosec}^{2} θ d θ = - \cot θ \end{cases}

I = \frac{- 1}{81} (\frac{\sqrt{{(9 - x)}^{3}}}{3 x^{3}} + \frac{\sqrt{9 - x^{2}}}{x}) + c

E x p l a n . \sin θ = \frac{x}{3} \Rightarrow a d j a c e n t = \sqrt{9 - x^{2}}

T h a n k s f o r a l l w h o s o l v e d

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