Calculate the value of $\int_0^\frac{\pi}{6} \frac{\cos x \operatorname d\!x}{\sqrt{\frac{1}{4}-\sin^2x}}$

Question

$$\int_0^\frac{\pi}{6} \frac{\cos x \operatorname d\!x}{\sqrt{\frac{1}{4}-\sin^2x}}$$

so $$\lim_{\epsilon->\frac{\pi}{6}} \int^{\epsilon} _{0} \frac{\cos x}{\sqrt{\frac{1}{4} - \sin^2x }} $$

$$\int \frac{\cos x}{\sqrt{\frac{1}{4} - \sin^2x }} = |t=\sin x, dt=\cos x|= \int \frac{dt}{\sqrt{\frac{1}{4}-t^2}}=\cdots=2\arcsin(2\sin x) $$

How can I prove this? What is the next step?

Answer 1

$$\int\frac{\cos x}{\sqrt{\frac14-\sin^2x}}dx=\int\frac{2\cos x\,dx}{\sqrt{1-\left(2\sin x\right)^2}}=\arcsin(2\sin x)+C$$

Using the fact that for a differentiable function $\;f\;$ we have that

$$\int \frac{ f'(x)}{\sqrt{1-f^2}}dx=\arcsin f(x)+C$$

Answer 2

Set $t=2\sin x$. Then $t\in [0,1]$, $dt=2\cos x\,dx$. Then $$ \int_0^\frac{\pi}{6} \frac{\cos x}{\sqrt{\frac{1}{4}-\sin^2x}}=\int_0^1\frac{dt}{\sqrt{1-t^2}}=\sin^{-1}t\,\big|_{\,t=0}^{\,t=1}=\frac{\pi}{2}. $$

Answer 3

The next step is to look up the derivative of $\arcsin$.

Answer 4

Now you want to use $\displaystyle\lim_{\epsilon\to {\frac{\pi}{6}}^{-}}\big[\arcsin(2\sin x)\big]_{0}^{\epsilon}=\lim_{\epsilon\to{\frac{\pi}{6}}^{-}}\big(\arcsin(2\sin \epsilon)-\arcsin(2\sin 0)\big)=\arcsin 1-\arcsin 0=\frac{\pi}{2}$