How do you use substitution to integrate $\frac{1}{\sqrt{x^{2} - 4}}$ $1/sqrt(x^2 -4)$ ?

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How do you use substitution to integrate $\frac{1}{\sqrt{x^{2} - 4}}$ $1/sqrt(x^2 -4)$ ?

1 Answer

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Answer 1 · 2015-06-23T17:39:02

ln I $\frac{x}{2} + \frac{\sqrt{x^{2} - 4}}{2}$ $x/2 +sqrt(x^2-4)/2$ I +C

Explanation:

Let x= 2sec u. so that dx= 2sec u tan u du. The given integral becomes $\int \frac{1}{\sqrt{4 {sec}^{2} u - 4}} 2 sec u tan u d u$ $int 1/sqrt (4sec^2 u -4) 2 sec u tan u du$

= $\int sec u d u$ $int secu du$

= ln lsecu +tan u I +C

= ln I $\frac{x}{2} + \frac{\sqrt{x^{2} - 4}}{2}$ $x/2 + sqrt(x^2-4)/2$ I +C

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How do you use substitution to integrate $\frac{1}{\sqrt{x^{2} - 4}}$ $1/sqrt(x^2 -4)$ ?

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Explanation:

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How do you use substitution to integrate 1/sqrt(x^2 -4)1√x2−4 1/sqrt(x^2 -4)?

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How do you use substitution to integrate $\frac{1}{\sqrt{x^{2} - 4}}$ $1/sqrt(x^2 -4)$ ?