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cmkluza

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## Homework Statement

Using the substitution x = 2sinθ, show that

[tex] \int \sqrt{4 - x^2} dx = Ax\sqrt{4 - x^2} + B ⋅ arcsin(\frac{x}{2}) + C [/tex]

whee A and B are constants whose values you are required to find.

## Homework Equations

## The Attempt at a Solution

x = 2sinθ

[itex] \frac{dx}{dθ} = 2cosθ [/itex]

dx = 2cosθ ⋅ dθ

[itex] \int \sqrt{4 - x^2} dx = \int \sqrt{4 - 4sin^2θ} ⋅ 2cosθ dθ [/itex]

Edit: (Hopefully) corrected my work after this point

[itex] 2\int \sqrt{4(1 - sin^2θ)} ⋅ cosθ dθ [/itex]

[itex] 2\int 2\sqrt{1 - sin^2θ} ⋅ cosθ dθ \rightarrow 1 - sin^2θ = cos^2θ [/itex]

[itex] 4\int \sqrt{cos^2θ} ⋅ cosθ dθ [/itex]

[itex] 4 \int cosθ ⋅ cosθ ⋅ dθ [/itex]

[itex] 4 \int cos^2θ ⋅ dθ [/itex]

[itex] 4 \int \frac{1 + cos2θ}{2} ⋅ dθ = 2 \int 1 + cos2θ ⋅ dθ [/itex]

2(θ + 2sin2θ)

x = 2sinθ ∴ θ = arcsin[itex]\frac{x}{2}[/itex]

2arcsin[itex]\frac{x}{2}[/itex] + 4sin2θI'm getting closer at this point, but can't figure out how to transform the 4sin2θ back into something to do with x's. If I substituted 2sinθcosθ in for sin2θ then I would be stuck with a cosine in the final equation that I don't want, nor would that get me any closer to a square root shown in the final answer. Have I messed up in my work up to this point again, or is there something I'm not looking at? Thanks!

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