Problem source

Let $y=\mathrm{f}(x)$, $(0\leqslant x\leqslant a)$, be a continuous curve lying in the first quadrant and passing through the origin.
Suppose that, for each non-negative value of $y$ with $0\leqslant y\leqslant\mathrm{f}(a)$, there is \textit{exactly} one value of $x$ such that $\mathrm{f}(x)=y$; thus we may write $x=\mathrm{g}(y)$, for a suitable function $\mathrm{g}.$ 
For $0\leqslant s\leqslant a,$ $0\leqslant t\leqslant \mathrm{f}(a)$, define 
\[
\mathrm{F}(s)=\int_{0}^{s}\mathrm{f}(x)\,\mathrm{d}x,\qquad\mathrm{G}(t)=\int_{0}^{t}\mathrm{g}(y)\,\mathrm{d}y.
\]
By a geometrical argument, show that 
\[
\mathrm{F}(s)+\mathrm{G}(t)\geqslant st.\tag{*}
\]
When does equality occur in $(*)$?
Suppose that $y=\sin x$ and that the ranges of $x,y,s,t$ are restricted to $0\leqslant x\leqslant s\leqslant\frac{1}{2}\pi,$ $0\leqslant y\leqslant t\leqslant1$.
By considering $s$ such that the equality holds in $(*)$, show that
\[
\int_{0}^{t}\sin^{-1}y\,\mathrm{d}y=t\sin^{-1}t-\left(1-\cos(\sin^{-1}t)\right).
\]
Check this result by differentiating both sides with respect to $t$.

Solution source

\begin{center}
\begin{tikzpicture}[scale=2]
    \draw[->] (-.1, 0) -- (3, 0);
    \draw[->] (0,-.1) -- (0,3);

    \node at (3,0) [right] {$x$};
    \node at (0,3) [above] {$y$};
    \node at (3,3) [above] {$y = f(x)$};

    \draw[domain = 0:3, samples=180, variable = \x]  plot ({\x},{\x + 0.14*sin(360*\x) + 0.075 * sin(90*\x)}); 

    \draw (1,0) -- (1, {1 + 0.14 * sin(360) + 0.075 * sin(90)});
    \draw (0,{1.8 + 0.14 * sin(360*1.8) + 0.075 * sin(90*1.8)}) -- (1.8, {1.8 + 0.14 * sin(360*1.8) + 0.075 * sin(90*1.8)});
    \draw[dashed] (1,0) -- (1, {1.8 + 0.14 * sin(360*1.8) + 0.075 * sin(90*1.8)}) -- (1.8, {1.8 + 0.14 * sin(360*1.8) + 0.075 * sin(90*1.8)});

    \fill [blue, opacity = 0.1, domain = 0:1, samples=180, variable = \x]
      (0, 0)
      -- plot ({\x},{\x + 0.14*sin(360*\x) + 0.075 * sin(90*\x)})
      -- (1, 0)
      -- cycle;

    \fill [red, opacity = 0.1, domain = 0:1.8, samples=180, variable = \x]
      (0, 0)
      -- plot ({\x},{\x + 0.14*sin(360*\x) + 0.075 * sin(90*\x)})
      -- (0,{1.8 + 0.14 * sin(360*1.8) + 0.075 * sin(90*1.8)})
      -- cycle;
\end{tikzpicture}
\end{center}

The blue area is $F(s)$ the red area is $G(t)$, the dashed rectangle (which is a subset of the red and blue areas) has area $st$ therefore $F(s) + G(t) \geq st$. Equality holds if $f(s) = t$.

\begin{align*}
&& \int_0^t \sin^{-1} y \d y + \int_0^{\sin^{-1} t} \sin x \d x &= t \sin^{-1} t \\
\Rightarrow && \int_0^t \sin^{-1} y \d y &= t \sin^{-1} t - \left [ -\cos (x)  \right]_0^{\sin^{-1} t} \\
&&&= t \sin^{-1} t - (1- \cos (\sin^{-1} t))
\end{align*}

Let $y = t \sin^{-1} t - (1- \cos (\sin^{-1} t))$ then,

\begin{align*}
\frac{\d y}{\d t} &= \sin^{-1} t +t \frac{\d}{\d t} \l \sin^{-1} (t) \r - \sin ( \sin^{-1} t)  \frac{\d}{\d t} \l \sin^{-1} (t) \r \\
&= \sin^{-1} t
\end{align*}
as required