If $$(2+\sin x) \frac{\mathrm{d} y}{\mathrm{~d} x}+(y+1) \cos x=0$$ and $$y(0)=1$$, then $$y\left(\frac{\pi}{2}\right)$$ is

If $$A=\left[\begin{array}{cc}2 a & -3 b \\ 3 & 2\end{array}\right]$$ and $$A \cdot \operatorname{adj} A=A A^T$$, then $$2 a+3 b$$ is

$$\int\left(\frac{\tan \left(\frac{1}{x}\right)}{x}\right)^2 d x=$$

$$\int \frac{1}{(x+2)(1+x)^2} d x$$ has the value