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Can anyone tell me if the following of my integration is right or wrong?

[tex]

\int_{-2}^{2} \sqrt{4 - x^2} \ dx

[/tex]

let

[tex]x = 2 sin \theta[/tex]

then

[tex]\frac{dx}{d\theta}=2cos\theta, \ dx=2cos\theta \ d\theta[/tex]

also

[tex]4-x^2 = 4-(2sin\theta)^2

= 4 - 4sin^2\theta = 4(1-sin^2\theta)=4cos^2\theta

[/tex]

thus

[tex]

\sqrt{4-x^2}=\sqrt{4cos^2\theta}

[/tex]

[tex]\int_{-2}^{2} \sqrt{4 - x^2} \ dx[/tex]

[tex]= \int_{-2}^{2} \sqrt{4cos^2\theta} \ (2cos\theta \ d\theta)[/tex]

[tex]= \int_{-2}^{2} 2cos\theta \ (2cos\theta \ d\theta)[/tex]

[tex]= 4\int_{-2}^{2} cos^2\theta \ d\theta[/tex]

[tex]= 4\int_{-2}^{2} \frac{1+cos2\theta}{2} \ d\theta[/tex]

[tex]= \frac{4}{2} \int_{-2}^{2} (1+cos2\theta) \ d\theta[/tex]

let [tex]u=2\theta[/tex] then [tex]\frac{du}{d\theta}=2, \ d\theta = \frac{1}{2}du[/tex]

[tex]= 2 \int_{-2}^{2} (1+cos2\theta) \ d\theta[/tex]

[tex]= 2 \int_{-2}^{2} (1+cos \ u) \ (\frac{1}{2}du)[/tex]

[tex]= \frac{2}{2}\int_{-2}^{2} (1+cos \ u) \ du[/tex]

[tex]= \int_{-2}^{2} (1+cos \ u) \ du[/tex]

[tex]= u + sin \ u]^{2}_{-2}[/tex]

[tex]= 2\theta + sin2\theta]^{2}_{-2}[/tex]

[tex]= [2(2) + sin2(2)] - [2(-2) + sin2(-2)][/tex]

[tex]= (4 + sin4) - [-4 + sin(-4)][/tex]

[tex]= 4 + sin4 + 4 - sin(-4)[/tex]

[tex]= 8 + sin4 - sin(-4)[/tex]

[tex]= 8 + 0.07 - (-0.035)[/tex]

[tex]= 8 + 0.07 + 0.035[/tex]

[tex]= 8 + 0.07 + 0.035[/tex]

[tex]= 8.105[/tex]

Thenk you very much for evaluating

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# Homework Help: Evaluate this integral please

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