Match the following.

(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)$$\begin{aligned}& \frac{d y}{d x}=\frac{2}{x+y} \Rightarrow \frac{d x}{d y}-\frac{x}{2}=\frac{y}{2}\\ & \Rightarrow \quad x e^{-y / 2}=\frac{1}{2} \cdot \inty \cdote^{-y / 2} d y \\ & \Rightarrow \quad x+y+2=k e^{y / 2} \\ & \Rightarrow \quad k=3 \\ & \therefore \quad x+y+2=3 e^{y / 2} \\ & \end{aligned}$$[passing through (1,0) ]