# How do you integrate #int sqrt(x^2+1)# by trigonometric substitution?

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To integrate ∫√(x^2 + 1) dx using trigonometric substitution, let x = tan(θ).

Then, dx = sec^2(θ) dθ, and √(x^2 + 1) = √(tan^2(θ) + 1) = sec(θ).

So the integral becomes ∫sec^2(θ) * sec(θ) dθ.

This simplifies to ∫sec^3(θ) dθ.

Now, you can use the reduction formula for ∫sec^n(θ) dθ:

∫sec^n(θ) dθ = (1/(n-1)) * sec^(n-2)(θ) * tan(θ) + (n-2)/(n-1) * ∫sec^(n-2)(θ) dθ

For n = 3, this becomes:

∫sec^3(θ) dθ = (1/2) * sec(θ) * tan(θ) + (1/2) * ∫sec(θ) dθ

Now, integrate ∫sec(θ) dθ using the substitution u = tan(θ):

∫sec(θ) dθ = ∫(1/cos(θ)) dθ = ∫(1/(1 + u^2)) du

This integral can be solved using a standard integral formula or by partial fractions.

Finally, substitute back θ = tan^(-1)(x) into the result obtained above to get the final answer in terms of x.

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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.

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.

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.

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