Compute $\int_0^{\frac\pi2} \frac1{\sin(x+\frac\pi3)\sin(x+\frac\pi6)} dx$ Compute the following definite integral:
$$\int_0^{\frac\pi2} \frac1{\sin(x+\frac\pi3)\sin(x+\frac\pi6)} dx$$
I know that I can rewrite the denominator as shown below:
$$\int_0^{\frac\pi2} \frac1{\frac12\left(\cos(\frac\pi6) - \cos(2x+\frac\pi2)\right)} dx$$
$$\int_0^{\frac\pi2} \frac4{\sqrt{3} + 2\sin(2x)} dx$$
And then use the substitution
$$\sin(2x) = \frac{2\tan(x)}{1+\tan^2(x)}$$
$$dx=\frac1{1+\tan^2(x)}dt$$
which results in having to compute
$$\int_0^\infty \frac4{\sqrt{3}t^2 + 4t + \sqrt{3}} = 2\ln(3)$$
However, is there a simpler (or quicker) way to solve this integral? Thank you!

Here is some trick:
$\int_0^{\frac\pi2} \frac4{\sqrt{3} + 2\sin(2x)} dx$=$\int_0^{\frac\pi2} \frac4{\sqrt{3} + 4\sin(x)\cos(x)} dx$=$\int_0^{\frac\pi2} \frac{4\sqrt{3}\sec^2{x}}{\sec^2{x}(3 + 4\sqrt{3}\sin(x)\cos(x))} dx$=$\int_0^{\frac\pi2} \frac{4\sqrt{3}}{(3(1+\tan^2{x}) + 4\sqrt{3}\tan{x})} d(\tan{x})$
=$-\int_0^{\frac\pi2} \frac{4}{1-(2+\sqrt{3}\tan{x})^2} d(2+\sqrt{3}\tan{x})$.
Now, it becomes trivial integral, and the result is $2\log{3}$ as desired.
