$$\int \frac{\mathrm{e}^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] \mathrm{d} x, x > 0=$$

If $$ I=\int \frac{\sin x+\sin ^3 x}{\cos 2 x} d x=P \cos x+Q \log \left|\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}\right| $$ (where $$c$$ is a constant of integration), then values of $$\mathrm{P}$$ and $$\mathrm{Q}$$ are respectively

$$\int \frac{1}{\sin (x-a) \sin x} d x=$$

If $$f(x)=\sqrt{\tan x}$$ and $$g(x)=\sin x \cdot \cos x$$ then $$\int \frac{f(x)}{g(x)} \mathrm{d} x$$ is equal to (where $$C$$ is a constant of integration)