How do you find the integral of $\frac{1}{x^{2} \sqrt{25 - x^{2}}} d x$ $1/(x^2sqrt(25-x^2))dx$ ?

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How do you find the integral of $\frac{1}{x^{2} \sqrt{25 - x^{2}}} d x$ $1/(x^2sqrt(25-x^2))dx$ ?

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Answer 1 · 2015-03-01T16:35:34

The answer is: $- \frac{\sqrt{25 - x^{2}}}{25 x} + c$ $-sqrt(25-x^2)/(25x)+c$ .

We have make a substitution:

$x = 5 sin t \Rightarrow d x = 5 cos t d t$ $x=5sintrArrdx=5costdt$ ,

so:

$\int \frac{1}{x^{2} \sqrt{25 - x^{2}}} d x = \int \frac{1}{{(5 sin t)}^{2} \sqrt{25 - 25 {sin}^{2} t}} \cdot 5 cos t d t =$ $int1/(x^2sqrt(25-x^2))dx=int1/((5sint)^2sqrt(25-25sin^2t))*5costdt=$

$= \int \frac{5 cos t}{{(5 sin t)}^{2} \sqrt{25 (1 - {sin}^{2} t)}} d t =$ $=int(5cost)/((5sint)^2sqrt(25(1-sin^2t)))dt=$

$= \int \frac{5 cos t}{25 {sin}^{2} t \cdot 5 cos t} d t = \frac{1}{25} \int \frac{1}{{sin}^{2} t} d t =$ $=int(5cost)/(25sin^2t*5cost)dt=1/25int1/sin^2tdt=$

$= - \frac{1}{25} \int - \frac{1}{{sin}^{2} t} d t = - \frac{1}{25} cot t + c = - \frac{1}{25} \frac{cos t}{sin t} + c = (1)$ $=-1/25int-1/sin^2tdt=-1/25cott+c=-1/25cost/sint+c=(1)$

since

$sin t = \frac{x}{5}$ $sint=x/5$ and $cos t = \sqrt{1 - {sin}^{2} t} = \sqrt{1 - \frac{x^{2}}{25}} = \frac{\sqrt{25 - x^{2}}}{5}$ $cost=sqrt(1-sin^2t)=sqrt(1-x^2/25)=sqrt(25-x^2)/5$

than:

$(1) = - \frac{1}{25} \frac{\sqrt{25 - x^{2}}}{5} \cdot \frac{5}{x} + c = - \frac{\sqrt{25 - x^{2}}}{25 x} + c$ $(1)=-1/25sqrt(25-x^2)/5*5/x+c=-sqrt(25-x^2)/(25x)+c$ .

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How do you find the integral of $\frac{1}{x^{2} \sqrt{25 - x^{2}}} d x$ $1/(x^2sqrt(25-x^2))dx$ ?

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