dx-x-2-25-x-2-

Question Number 119970 by bramlexs22 last updated on 28/Oct/20

\int \frac{d x}{x^{2} \sqrt{25 - x^{2}}} ?

Answered by bemath last updated on 28/Oct/20

\int \frac{d x}{x^{3} \sqrt{\frac{25}{x^{2}} - 1}} d x = \int \frac{x^{- 3}}{\sqrt{25 x^{- 2} - 1}} d x

l e t t i n g m = 25 x^{- 2} - 1 \Rightarrow d m = - 50 x^{- 3} d x

\int \frac{- \frac{d m}{50}}{\sqrt{m}} = - \frac{1}{50} \int m^{- \frac{1}{2}} d m

= - \frac{1}{25} \sqrt{m} + c = - \frac{1}{25} \sqrt{25 x^{- 2} - 1} + c

= - \frac{\sqrt{25 - x^{2}}}{25 x} + c

Answered by Dwaipayan Shikari last updated on 28/Oct/20

\int \frac{d x}{25 {s i n}^{2} θ \sqrt{25 - 25 {s i n}^{2} θ}} x = 5 s i n θ \Rightarrow 1 = 5 c o s θ \frac{d θ}{d x}

= \int \frac{5 c o s θ d θ}{25 {s i n}^{2} θ . 5 c o s θ}

= \frac{1}{25} \int \frac{1}{{s i n}^{2} θ}

= - \frac{1}{25} {(t a n θ)}^{- 1} = - \frac{1}{25} {(. \frac{s i n θ}{\sqrt{1 - {s i n}^{2} θ}})}^{- 1} = - \frac{1}{25} {(. \frac{5 x}{5 \sqrt{25 - x^{2}}})}^{- 1} = - \frac{1}{25} . {(\frac{x}{\sqrt{25 - x^{2}}})}^{- 1} + C

= - \frac{1}{25} (\frac{\sqrt{25 - x^{2}}}{x}) + C

Commented by bemath last updated on 28/Oct/20

\cot θ = \frac{\cos θ}{\sqrt{1 - \cos^{2} θ}}

Commented by Dwaipayan Shikari last updated on 28/Oct/20

I s u p p o s e i t a s t a n θ t y p o

T h a n k i n g f o r c o r r e c t i o n

Answered by Bird last updated on 28/Oct/20

I = \int x^{- 2} {(25 - x^{2})}^{- \frac{1}{2}} d x b y p a r t s

I = - \frac{1}{x} {(25 - x^{2})}^{- \frac{1}{2}} - \int - \frac{1}{x} (- \frac{1}{2}) (- 2 x) {(25 - x^{2})}^{- \frac{3}{2}}

= - \frac{1}{x \sqrt{25 - x^{2}}} + \int \frac{d x}{(25 - x^{2}) \sqrt{25 - x^{2}}}

\int \frac{d x}{(25 - x^{2}) \sqrt{25 - x^{2}}} =_{x = 5 s i n t}

= \int \frac{5 c o s t d t}{25 {c o s}^{2} t . 5 . c o s t}

= \frac{1}{25} \int \frac{d t}{\frac{1 + c o s (2 t)}{2}} = \frac{2}{25} \int \frac{d t}{1 + c o s (2 t)}

=_{t a n t = u} \frac{2}{25} \int \frac{d u}{(1 + u^{2}) (1 + \frac{1 - u^{2}}{1 + u^{2}})}

= \frac{2}{25} \int \frac{d u}{1 + u^{2} + 1 - u^{2}} = \frac{1}{25} \int d u

= \frac{u}{25} + c = t a n (a r c s i n (\frac{x}{5})) + c \Rightarrow

I = - \frac{1}{x \sqrt{25 - x^{2}}} + t a n (a r c s i n (\frac{x}{5}) + C

Commented by Bird last updated on 28/Oct/20

I = - \frac{1}{x \sqrt{25 - x^{2}}} + \frac{1}{25} a r c t a n (a r c s i n (\frac{x}{5})) + C

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