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

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To integrate ∫ dx / sqrt(x^2 + 4) using trigonometric substitution, let x = 2tan(θ), then dx = 2sec^2(θ) dθ. Substitute these into the integral:

∫ (2sec^2(θ) dθ) / sqrt((2tan(θ))^2 + 4)

Simplify the expression under the square root:

∫ (2sec^2(θ) dθ) / sqrt(4tan^2(θ) + 4) = ∫ (2sec^2(θ) dθ) / sqrt(4(sec^2(θ)) + 4) = ∫ (2sec^2(θ) dθ) / sqrt(4sec^2(θ) + 4)

Combine like terms under the square root:

= ∫ (2sec^2(θ) dθ) / sqrt(4(sec^2(θ) + 1))

Now, using the identity sec^2(θ) + 1 = tan^2(θ) + 1 = sec^2(θ), we simplify further:

= ∫ (2sec^2(θ) dθ) / (2sec(θ)) = ∫ sec(θ) dθ

The integral of sec(θ) with respect to θ is ln|sec(θ) + tan(θ)| + C.

Finally, substitute back for θ using the original substitution x = 2tan(θ):

ln|sec(θ) + tan(θ)| + C = ln|sec(arctan(x/2)) + tan(arctan(x/2))| + C = ln|√(x^2 + 4) + x/2| + C

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To integrate ( \int \frac{dx}{\sqrt{x^2 + 4}} ) using trigonometric substitution, we make the substitution ( x = 2\tan(\theta) ).

Then, ( dx = 2\sec^2(\theta) , d\theta ), and ( \sqrt{x^2 + 4} = \sqrt{4\tan^2(\theta) + 4} = 2\sec(\theta) ).

Substituting these into the integral, we get:

[ \int \frac{2\sec^2(\theta) , d\theta}{2\sec(\theta)} ]

[ = \int \sec(\theta) , d\theta ]

The integral of ( \sec(\theta) ) can be evaluated using the formula ( \int \sec(\theta) , d\theta = \ln|\sec(\theta) + \tan(\theta)| + C ).

Thus, the integral ( \int \frac{dx}{\sqrt{x^2+4}} ) can be expressed as ( \ln|\sec(\theta) + \tan(\theta)| + C ), where ( C ) is the constant of integration.

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