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How do you evaluate the integral #int dx/sqrt(a^2+x^2)#?

Answer 1

Evelyn Agard

#= sinh^(-1) (x/a) + C#

Because #cosh^2 z - sinh^2 z = 1#, I would go with the hyperbolic sub #x = a sinh y, dx = a cosh y \ dy#

The integral becomes:

#int 1/sqrt(a^2+a^2 sinh^2 y)a cosh y \ dy#

#= int 1/(a cosh y)*a cosh y \ dy#

#= y + C#

#= sinh^(-1) (x/a) + C#

And because #sinh^(-1) z = ln ( z + sqrt(z^2 + 1))#

#= ln ( x/a + sqrt((x/a)^2 + 1)) + C#

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Answer 2

Ezra Adams

To evaluate the integral ∫ dx/√(a^2+x^2), where a is a constant, you can use a trigonometric substitution. Let x = a * tan(θ), then dx = a * sec^2(θ) dθ. Substitute these into the integral and simplify, leading to the integral ∫ (a * sec^2(θ))/√(a^2+a^2tan^2(θ)) dθ. Simplify further to get ∫ (a * sec^2(θ))/√(a^2(1+tan^2(θ))) dθ, which equals ∫ (a * sec^2(θ))/√(a^2*sec^2(θ)) dθ. Cancel out the common terms to obtain ∫ dθ. This integral evaluates to θ + C. Finally, substitute back θ = arctan(x/a) into the result to get the final answer: arctan(x/a) + C.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.