(i) $I=\int_0^{\pi / 2}(\sin x)^{\cos x}\left(\cos x \cot x-\log (\sin x)^{\sin x}\right) d x=\int_0^{\pi / 2} \frac{d}{d x}(\sin x)^{\cos x} d x=1$
(ii) The point of intersection of $-4 y^2=x$ and $x-1=-5 y^2$ is $(-4,-1)$ and $(-4,1)$.
Hence, required area $=2\left\{\int_0^1\left(1-5 y^2\right) d y-\int_0^1-4 y^2 d y\right\}=\frac{4}{3}$.
(iii) The point intersection of $y=3^{x-1} \log x$ and $y=x^x-1$ is $(1,0)$.
Hence, $\quad \frac{d y}{d x}=\frac{3^{x-1}}{x}+3^{x-1} \cdot \log 3 \cdot \log x$
$\therefore \quad\left(\frac{d y}{d x}\right)_{(1,0)}=1$
For
$$
\therefore \quad\left(\frac{d y}{d x}\right)_{(1,0)}=1
$$

If $\theta$ is angle between the curves, then $\tan \theta=0$.
$$
\because \quad \theta=0^{\circ}
$$
(iv)
$$
\begin{aligned}
& \frac{d y}{d x}=\frac{2}{x+y} \Rightarrow \frac{d x}{d y}-\frac{x}{2}=\frac{y}{2} \\
& \Rightarrow \quad x e^{-y / 2}=\frac{1}{2} \cdot \int y \cdot e^{-y / 2} d y \\
& \Rightarrow \quad x+y+2=k e^{y / 2} \\
& \Rightarrow \quad k=3 \\
& \therefore \quad x+y+2=3 e^{y / 2} \\
&
\end{aligned}
$$
[passing through $(1,0)$ ]