How do you integrate $\int \frac{d x}{\sqrt{x^{2} + 4}}$ $int dx/sqrt(x^2+4)$ using trig substitutions?

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How do you integrate $\int \frac{d x}{\sqrt{x^{2} + 4}}$ $int dx/sqrt(x^2+4)$ using trig substitutions?

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Answer 1 · 2016-12-14T15:56:16

Substitute $x = 2 sinh u$ $x=2 sinh u$ quickly getting the integral to be ${sinh}^{- 1} (\frac{x}{2}) + C$ $sinh^{-1}(x/2)+C$

Explanation:

$x = 2 sinh u$ $x=2sinh u$ , $d x = 2 cosh u d u$ $dx=2cosh u du$ giving the integral as
$\int 2 \frac{cosh u}{\sqrt{4 {sinh}^{2} u + 4}} d u$ $int2coshu/ sqrt(4 sinh^2u+4)du$
$= \int \frac{cosh u}{cosh u} d u$ $=int cosh u/ cosh u du$ because ${cosh}^{2} u - {sinh}^{2} u = 1$ $cosh^2u -sinh^2 u =1$
$= \int 1 d u$ $=int1du$
$= u + C$ $=u+C$
$= {sinh}^{- 1} (\frac{x}{2}) + C$ $=sinh^-1(x/2)+C$

Some people prefer to write this as $ln (x + \sqrt{x^{2} + 4})$ $ln(x+sqrt(x^2+4))$ which is equivalent, as it differs from the first solution by a fixed constant $ln (\frac{1}{2})$ $ln(1/2)$ .

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How do you integrate $\int \frac{d x}{\sqrt{x^{2} + 4}}$ $int dx/sqrt(x^2+4)$ using trig substitutions?

1 Answer

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How do you integrate ∫dx√x2+4int dx/sqrt(x^2+4) using trig substitutions?

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Explanation:

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How do you integrate $\int \frac{d x}{\sqrt{x^{2} + 4}}$ $int dx/sqrt(x^2+4)$ using trig substitutions?