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

The way to approach these types of problems is try and make a comparison with a known trig identity. In the case we will use:

If you compare to the integrand, we make the following substitution:

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To integrate ∫√(1 - x^2) dx using trigonometric substitution, let x = sin(θ). Then, dx = cos(θ) dθ. Substituting these into the integral gives:

∫√(1 - sin^2(θ)) * cos(θ) dθ

Simplify the expression under the square root:

√(1 - sin^2(θ)) = √(cos^2(θ)) = cos(θ)

So, the integral becomes:

∫cos^2(θ) dθ

Now, use the trigonometric identity cos^2(θ) = (1 + cos(2θ))/2:

∫(1 + cos(2θ))/2 dθ

Integrate term by term:

∫(1/2 + cos(2θ)/2) dθ = (1/2)∫dθ + (1/2)∫cos(2θ) dθ

Integrate each term:

(1/2)θ + (1/4)sin(2θ) + C

Now, substitute back for θ:

(1/2)sin^(-1)(x) + (1/4)sin(2* sin^(-1)(x)) + C

Using the double angle identity sin(2θ) = 2sin(θ)cos(θ), we get:

(1/2)sin^(-1)(x) + (1/4)sin(sin^(-1)(x))cos(sin^(-1)(x)) + C

Remembering that sin^(-1)(x) is the same as arcsin(x), the final answer is:

(1/2)sin^(-1)(x) + (1/4)x√(1 - x^2) + C

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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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