$$\begin{align*}I & = 2\int_{0}^{\frac{1}{\sqrt{2}}} \dfrac{\sin^{-1} x}{x}\ dx - \int_{0}^{1} \dfrac{\tan^{-1} x}{x} \ dx \\ & = 2\int_{0}^{\frac{\pi}{4}} \dfrac{\theta \cos \theta}{\sin \theta} \ d\theta - \int_{0}^{\frac{\pi}{4}} \dfrac{\theta \sec^2 \theta}{\tan \theta} \ d\theta \\ & = \color{red}{-\dfrac{\pi}{4} \ln 2} -2\int_{0}^{\frac{\pi}{4}} \ln \sin \theta \ d\theta + \int_{0}^{\frac{\pi}{4}} \ln \tan \theta \ d\theta \\ & = -\int_{0}^{\frac{\pi}{4}} \ln \sin 2\theta \ d\theta \\ & = \dfrac{\pi}{4} \ln 2\end{align*}$$
Could someone tell me how the author of this book (Vikas Gupta, Advanced Problems in Mathematics) came up with the 3rd step containing log? Is it some sort of standard result?
What makes me ask is cause of the term $-\frac\pi4 \log 2$
If I understood well, your question is about
$$2\int_{0}^{\frac{\pi}{4}} \dfrac{\theta \cos \theta}{\sin \theta} \ d\theta - \int_{0}^{\frac{\pi}{4}} \dfrac{\theta \sec^2 \theta}{\tan \theta} \ d\theta$$
Turning into $$-\dfrac{\pi}{4} \ln 2 -2\int_{0}^{\frac{\pi}{4}} \ln \sin \theta \ d\theta + \int_{0}^{\frac{\pi}{4}} \ln \tan \theta \ d\theta$$
From integrating by parts:
$$\int u \ dv = u \cdot v - \int v \ du$$
Therefore
$$\begin{align*}\int_{0}^{\pi/4} \underbrace{\theta}_{u} \cdot \underbrace{\dfrac{\cos \theta}{\sin \theta} \ d\theta}_{dv} & = \left[\theta \cdot \ln \left(\sin \theta\right)\right]_{0}^{\pi/4} - \int_{0}^{\pi/4}\ln \left(\sin \theta\right)\ d\theta \\ & = \dfrac{\pi}{4}\cdot \ln \left(\sin \dfrac{\pi}{4}\right) - \underbrace{0 \cdot \ln \left(\sin 0\right)}_{\text{careful}} - \int_{0}^{\pi/4}\ln \left(\sin \theta\right)\ d\theta \end{align*} $$
The limit which I put careful is $0$ from wolfram alpha.
Now, the log term is:
$$\begin{align}\dfrac{\pi}{4}\cdot \ln \left(\sin \dfrac{\pi}{4}\right) & = \dfrac{\pi}{4}\cdot \ln \dfrac{1}{\sqrt{2}} \\ & = \dfrac{\pi}{4}\left(-\ln \sqrt{2}\right) \\ & = \dfrac{-\pi}{4} \ln \sqrt{2} \\ & = \dfrac{-\pi}{4} \cdot \dfrac{1}{2} \ln 2 \\ & = \dfrac{-\pi}{8} \cdot \ln 2\end{align}$$
Therefore:
$$\boxed{2\int_{0}^{\frac{\pi}{4}} \dfrac{\theta \cos \theta}{\sin \theta} \ d\theta = 2\cdot \dfrac{-\pi}{8} \cdot \ln 2 - 2\int_{0}^{\pi/4}\ln \left(\sin \theta\right)\ d\theta}$$