Hot integration 112

Similar solution as @Tom Engelsman

$\begin{aligned} I & = \int_0^{\sin^{-1}\frac 13} \frac {\cos x}{2\cos (2x)-1} dx \\ & = \int_0^{\sin^{-1}\frac 13} \frac {\cos x}{2(1-2\sin^2 x)-1} dx \\ & = \int_0^{\sin^{-1}\frac 13} \frac {\cos x}{1-4\sin^2 x} dx & \small \blue{\text{Let }u = \sin x \implies du = \cos x \ dx} \\ & = \int_0^\frac 13 \frac 1{1-4u^2} du \\ & = \int_0^\frac 13 \frac 12 \left(\frac 1{1-2u} + \frac 1{1+2u} \right) du \\ & = \frac 12 \left[-\frac {\ln (1-2u)}2 + \frac {\ln (1+2u)}2 \right]_0^\frac 13 \\ & = \frac 14 \ln \left(\frac {1+2u}{1-2u} \right) \bigg|_0^\frac 13 \\ & = \frac {\ln 5}4 \approx \boxed{0.402} \end{aligned}$

Let us rewrite the above integrand as the following:

$\frac{\cos(x)}{2\cos(2x)-1} = \frac{\cos(x)}{2(2\cos^{2}(x)-1)-1} = \frac{\cos(x)}{4\cos^{2}(x)-3} = \frac{\cos(x)}{4(1-\sin^{2}(x))-3} = \frac{\cos(x)}{1-4\sin^{2}(x)} = \frac{1}{2} \cdot [\frac{1}{1+2\sin(x)} + \frac{1}{1-2\sin(x)}] \cdot \cos(x)$ .

Let $u = \sin(x), du = \cos(x) dx, 0 \le u \le \frac{1}{3}$ , which now results in:

$\frac{1}{2} \int_{0}^{\frac{1}{3}} \frac{1}{1+2u} + \frac{1}{1-2u} du;$

or $\frac{1}{4} \cdot [\ln(1+2u) - \ln(1-2u)] |_{0}^{\frac{1}{3}}$ ;

or $\boxed{\frac{1}{4} \ln(5)}.$