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

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

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

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

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

• 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. Commented May 24, 2015 at 23:18

Let $u=4x \Rightarrow du=4dx$

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

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