Evaluate this definite integral $$\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{-\ln(\tan\left(\frac{x}{2}\right))\cos^4\left(\frac{x}{2}\right)}dx$$
I used Weierstrass substitution and then Feynman's integration technique (Differentiation under the integral sign) and then got the answer $2\ln(3)$ and I'm interested to see if there are any faster or alternate methods to evaluate this integral.
Edit: I wrote $\tan\frac{\pi}{2}$ originally instead of $\tan\frac{x}{2}$ and it's now corrected
On
Hint: \begin{align} I&=\int_{0}^{\frac{\pi}{2}}\frac{\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})}{-\ln(\tan\left(\frac{x}{2}\right))\cos^4\left(\frac{x}{2}\right)}dx\\ &=2\int_{0}^{\frac{\pi}{2}} \frac{1-\tan^2(\frac{x}{2})}{-\ln(\tan\left(\frac{x}{2}\right))} d(\tan\frac{x}{2})\\ &=2\int_0^1 \frac{1-x^2}{-\ln x} dx \\ &=2\int_{0}^{\infty} \frac{e^{-t}-e^{-3t}}{t} dt \\ &=2\ln 3(Frullani \ integral) \end{align}
On
Substitute $y=\cot\dfrac x2$, then introduce a parameterized integral to be evaluated the Feynman way:
$$\begin{align*} &\int_0^{\tfrac\pi2}\frac{\cos x}{-\ln \left(\tan \frac x2\right) \, \cos^4\frac x2}\, dx \\ &= 4 \int_0^{\tfrac\pi2} \frac{\cos x}{(1+\cos x)^2 \ln \left(\cot \frac x2\right)} \, dx \\ &= 8 \int_1^\infty \frac{\frac{y^2-1}{y^2+1}}{\left(1+\frac{y^2-1}{y^2+1}\right)^2 \ln y} \, \frac{dy}{y^2+1} \\ &= 2 \int_1^\infty \frac{y^2-1}{y^4 \ln y} \, dy = 2I(2) \\[2ex] I(a) &= \int_1^\infty \frac{y^a-1}{y^4 \ln y} \, dy & (I(0)=0) \\[1ex] I'(a) &= \int_1^\infty y^{a-4} \, dy = \frac1{3-a} & (a<3) \\[1ex] \implies I(a) &= \ln3 - \ln(3-a) \end{align*}$$
Hence
$$\int_0^{\tfrac\pi2} \frac{\cos x}{-\ln\left(\tan\frac x2\right) \, \cos^4\frac x2} \,dx = \boxed{2\ln3}$$
Substitute $t= -\ln(\tan\frac{x}{2})$ to replace the inconvenient log function in the denominator
\begin{align} I= \int_0^{\frac{\pi}{2}}\frac{\cos x}{-\ln(\tan\frac{x}{2})\cos^4\frac{x}{2}}dx =2\int_0^\infty \frac{e^{-t} -e^{-3t}}t dt \end{align}
Then, integrate by parts
\begin{align} I &= 2\ln t(e^{-t} -e^{-3t})|_0^\infty -2\int_0^\infty\ln t(-e^{-t} +3e^{-3t})dt \\ & \overset {u=3t} = 2\int_0^\infty\ln t \> e^{-t}dt - 2\int_0^\infty (\ln u -\ln3) \>e^{-u}du \\ &= 2\ln3 \int_0^\infty e^{-u}du= 2\ln3 \end{align}