$$\text { If } y=\left(\sin ^{-1} x\right)^2+\left(\cos ^{-1} x\right)^2, \text { then }\left(1-x^2\right) y_2-x y_1=$$

If $$a>0$$ and $$z=\frac{(1+i)^2}{a-i}, i=\sqrt{-1}$$, has magnitude $$\frac{2}{\sqrt{5}}$$, then $$\bar{z}$$ is

In $$\triangle \mathrm{ABC}$$, with usual notations, $$2 \mathrm{ac} \sin \left(\frac{1}{2}(\mathrm{~A}-\mathrm{B}+\mathrm{C})\right)$$ is equal to

$$\int \frac{\sin 2 x\left(1-\frac{3}{2} \cos x\right)}{e^{\sin ^2 x+\cos ^3 x}} d x=$$