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

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

1 Answer

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Answer 1 · 2015-09-09T07:50:19

$\sqrt{1 - \frac{1}{x^{2}}} + C$ $sqrt(1-1/x^2) + C$

Explanation:

Finding this integral is easy by susbstituting $x = sec θ$ $x= sectheta$ so that $d x = sec θ tan θ d θ$ $dx= sectheta tantheta d theta$

$\int \frac{d x}{x^{2} \cdot \sqrt{x^{2} - 1}}$ $int dx/(x^2* sqrt (x^2-1)$ = $\int \frac{sec θ tan θ d θ}{{sec}^{2} θ tan θ}$ $int (sectheta tan theta d theta)/(sec^2 theta tan theta$

= $\int cos θ d θ$ $int costheta d theta$ = $sin θ + C$ $sintheta +C$ = $\sqrt{1 - \frac{1}{x^{2}}} + C$ $sqrt(1-1/x^2) + C$

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

1 Answer

Explanation:

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