Just another s $\pi$ cy integral

Refer @N. Aadhaar Murty solution. I will only give you a way to find $\displaystyle\int_0^{π/2}\log(\cos(x))dx$

Suppose $\int_0^{π/2} \log(\cos(x)) dx= \tau'(0)$ Now $\displaystyle \int_0^{π/2} \left(\cos(x)\right )^adx= \dfrac{\Gamma{(\frac{a+1}{2})}\Gamma{(\frac{1}{2})}}{2\Gamma{(\frac{a+2}{2})}}=\tau(a)$

Note From definition of Beta function we get $\displaystyle\int_0^{π/2} \sin^{2a-1} (x) \cos^{2b-1} (x) dx= \dfrac{\Gamma{(a)} \Gamma{(b)}}{2\Gamma{(a+b)}}$

Now $\displaystyle\int_0^{π/2} \log(\cos(x))\cos^a(x) dx=\dfrac{\partial}{\partial{a}} \left(\dfrac{\Gamma{(\frac{a+1}{2})}\Gamma{(\frac{1}{2})}}{2\Gamma{(\frac{a+2}{2})}}\right)$

$=\frac{\sqrt{\pi}}{\mathrm{4}}\left(\:\frac{\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\Gamma'\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)-\Gamma'\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)}\right)$ $=\frac{\sqrt{\pi}}{\mathrm{4}}\left(\frac{\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\psi\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)-\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\psi\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)}\:\right)=\tau'\left({a}\right)$ Now $\tau'\left(\mathrm{0}\right)=\frac{\sqrt{\pi}}{\mathrm{4}}\left(\frac{\Gamma\left(\mathrm{1}\right)\sqrt{\pi}\:\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)-\psi\left(\mathrm{1}\right)\sqrt{\pi}}{\mathrm{1}}\right)=\frac{\pi}{\mathrm{4}}\left(-\gamma-\mathrm{2log}\left(\mathrm{2}\right)+\gamma\right)$ $=-\frac{\pi}{\mathrm{2}}\mathrm{log}\left(\mathrm{2}\right)$ Similar approach for finding $\displaystyle\int_0^{π/2} \log^n(\sin(x))\sin^a(x)dx$

Using the definition for $\text {tanh}^{-1},$

$\text {tanh}^{-1} (\tan^2 x) = \frac {1}{2}\ln\left(\frac {\tan^2 x +1}{1 - \tan^2 x}\right) = \frac {1}{2}\ln\left(\frac {\sec^2 x}{-(\sec^2 x - 2)}\right) = -\frac {1}{2}\ln\left(\frac {-(\sec^2 x -1) + 1}{\sec^2 x}\right) = -\frac {1}{2}\ln\left(\frac {-(\tan^2 x) + 1}{\sec^2 x}\right) = -\frac {1}{2}\ln\left(\cos^2 x - \sin^2 x\right) = -\frac {1}{2}\ln\left(\cos(2x)\right)$

Therefore, our integral is

$\frac {-1}{2}\int_{0}^{\frac {1}{2}\pi} \ln(\cos x) dx$

The above integral has the well-known value of $-\frac {\pi}{2}\ln(2)$ which can be evaluated like this . Which makes the answer $\frac {\pi}{4}\ln(2) \approx \boxed {0.544}$ .