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integration equation

$$\int_{0}^{1/8} \frac{4}{\sqrt{(1-4x^2)}} \,dx$$

my work

$t= \sqrt{(1-4x^2)} $

$dt = -4x/\sqrt{(1-4x^2)} dx $

stuck here also

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3 Answers 3

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Use the substitution $2x=\sin \theta$. Then $\frac{d}{d\theta}x=\frac{1}{2}\cos \theta $ and the integral becomes $$\int_{0}^{1/8} \frac{4}{\sqrt{(1-4x^2)}} \,dx = \int_{0}^{\arcsin\left(\frac{1}{4}\right)} 2\,d\theta$$

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You can use integration by substitution $ 2x=sin(u) $. Then $ \frac{d}{du}x=\frac{1}{2}\cos(u) $.

We can rearrange our substitution equation $ 2x=sin(u) $ into $ u = arcsin(2x) $. So we can find our limits with respect to u. When $ x = \frac{1}{8} \: \: u = arcsin(2\frac{1}{8}) = arcsin(\frac{1}{4}) $ and when $ x = 0 \: \: u = arcsin(0) = 0 $. So

$ \int_{0}^{1/8} \frac{4}{\sqrt{(1-4x^2)}} \,dx = \int_{0}^{\arcsin\left(\frac{1}{4}\right)} 2\,du = 2\arcsin\left(\frac{1}{4}\right) = 0.50536051 $

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  • $\begingroup$ Upvoted. Maybe you could add a clarification on why the new integration limits are $0$ and $\arcsin\left(\frac{1}{4}\right)$, because it might not be clear for the asker. $\endgroup$
    – Pedro A
    Commented May 24, 2015 at 23:18
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Let $u=4x \Rightarrow du=4dx$

Therefore $\int\frac{4}{\sqrt{1-4x^2}}dx=8\int\frac{1}{\sqrt{4-u^2}}=8\arcsin(2x)+C$

$\Rightarrow\int_{0}^{1/8}\frac{4}{\sqrt{(1-4x^2)}}dx=8\arcsin(\frac{1}{4})$

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