How do you evaluate the integral $\int \sqrt{1 - \frac{1}{x^{2}}}$ $int sqrt(1-1/x^2)$ ?

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How do you evaluate the integral $\int \sqrt{1 - \frac{1}{x^{2}}}$ $int sqrt(1-1/x^2)$ ?

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Answer 1 · 2017-06-24T21:26:11

$I = \int \sqrt{1 - \frac{1}{x^{2}}} d x$ $I=intsqrt(1-1/x^2)dx$

Let's try to get the integrand to resemble $\sqrt{1 - {sin}^{2} θ}$ $sqrt(1-sin^2theta)$ . To do so, let $x = csc θ$ $x=csctheta$ . Then, $d x = - csc θ cot θ d θ$ $dx=-cscthetacotthetad theta$ , and:

$I = \int \sqrt{1 - \frac{1}{{csc}^{2} θ}} (- csc θ cot θ d θ)$ $I=intsqrt(1-1/csc^2theta)(-cscthetacotthetad theta)$

$I = \int \sqrt{1 - {sin}^{2} θ} (- csc θ cot θ) d θ$ $color(white)I=intsqrt(1-sin^2theta)(-cscthetacottheta)d theta$

$I = - \int cos θ \frac{1}{sin θ} \frac{cos θ}{sin θ} d θ$ $color(white)I=-intcostheta1/sinthetacostheta/sinthetad theta$

$I = - \int {cot}^{2} θ d θ$ $color(white)I=-intcot^2thetad theta$

Use ${cot}^{2} θ = {csc}^{2} θ - 1$ $cot^2theta=csc^2theta-1$ :

$I = \int (1 - {csc}^{2} θ) d θ$ $I=int(1-csc^2theta)d theta$

$I = θ + cot θ$ $color(white)I=theta+cottheta$

Reverse the substitution $x = csc θ$ $x=csctheta$ :

$I = θ + \sqrt{{csc}^{2} θ - 1}$ $I=theta+sqrt(csc^2theta-1)$

$I = {csc}^{- 1} x + \sqrt{x^{2} - 1} + C$ $color(white)I=csc^-1x+sqrt(x^2-1)+C$

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How do you evaluate the integral $\int \sqrt{1 - \frac{1}{x^{2}}}$ $int sqrt(1-1/x^2)$ ?

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