$$\int \frac{x+\sin x}{1+\cos x} d x=$$

If $$\int \frac{\sin \theta}{\sin 3 \theta} d \theta=\frac{1}{2 k} \log \left|\frac{k+\tan \theta}{k-\tan \theta}\right|+c$$, then $$k=$$

If $$\int \sqrt{x-\frac{1}{x}}\left(\frac{x^2+1}{x^2}\right) d x=\frac{2}{3}\left(x-\frac{1}{x}\right)^k+c$$, then value of $$k$$ is

$$\int \cot x \cdot \log [\log (\sin x)] d x=$$