Consider the following two curves
$$x ^ { 2 } + y ^ { 2 } = 1 , \tag{1}$$ $$4 x y = 1 , \tag{2}$$
where $x > 0 , y > 0$. We are to find the area $S$ of the region bounded by curve (1) and curve (2).
(1) First, let P and Q be the intersection points of curves (1) and (2), and let us denote the $x$-coordinates of P and Q by $p$ and $q$ $(p < q)$, respectively.
From (1), the coordinates $( x , y )$ of the intersection points of curves (1) and (2) can be expressed as $x = \cos \theta , y = \sin \theta \left( 0 < \theta < \frac { \pi } { 2 } \right)$. Then from (2) we have
$$\sin \mathbf { A } \theta = \frac { \mathbf { B } } { \mathbf{C} } .$$
From this we know that
$$\theta = \frac { \mathbf { D } } { \mathbf { E F } } \pi \quad \text { or } \quad \frac { \mathbf { G } } { \mathbf { H I } } \pi$$
(Write the answers in the order such that $\frac { \mathbf{D} } { \mathbf{EF} } < \frac { \mathbf{G} } { \mathbf{HI} }$.) Hence we have
$$p = \cos \frac { \mathbf { J } } { \mathbf { KL } } \pi , \quad q = \cos \frac { \mathbf { M } } { \mathbf { N } } \pi .$$
(2) Now we can find the value of $S$. Since
$$S = \int _ { p } ^ { q } \left( \sqrt { 1 - x ^ { 2 } } - \frac { 1 } { 4 x } \right) d x$$
we have to find the values of
$$I = \int _ { p } ^ { q } \sqrt { 1 - x ^ { 2 } } \, d x , \quad J = \int _ { p } ^ { q } \frac { 1 } { x } \, d x$$
For $I$, when we set $x = \cos \theta$ and calculate by substituting it for $x$ in the integral, we have
$$I = \frac { \mathbf { P } } { \mathbf { Q } }$$
For $J$, we have
$$J = \log \left( \mathbf { R } ^ { \mathbf{S} } + \sqrt { \mathbf { S } } \right) ,$$
where $\log$ is the natural logarithm. From these, we obtain
$$S = \frac { \mathbf{P} } { \mathbf{Q} } \pi - \frac { \mathbf { T } } { \mathbf{U} } \log ( \mathbf { R } + \sqrt { \mathbf { S } } ) .$$

Let $k$ be a positive real number. Consider the two curves
$$C _ { 1 } : y = \sin ^ { 2 } x , \quad C _ { 2 } : y = k \cos 2 x \quad \left( 0 \leqq x \leqq \frac { \pi } { 2 } \right)$$
Let $S _ { 1 }$ be the area of the region bounded by the two curves $C _ { 1 } , C _ { 2 }$ and the $y$-axis, and let $S _ { 2 }$ be the area of the region bounded by the two curves $C _ { 1 } , C _ { 2 }$ and the straight line $x = \frac { \pi } { 2 }$. We are to show that the value of $S _ { 2 } - S _ { 1 }$ is a constant independent of the value of $k$.
When we denote the $x$ satisfying the equation $\sin ^ { 2 } x = k \cos 2 x$ by $\alpha$, we have
$$\sin \alpha = \sqrt { \frac { k } { \mathbf { A } k + \mathbf { B } } } , \quad \cos \alpha = \sqrt { \frac { k + \mathbf { C } } { \mathbf { D } k + \mathbf { E } } } .$$
Then we have
$$\begin{aligned} & S _ { 1 } = \frac { \mathbf { F } } { \mathbf { F G } } \int _ { 0 } ^ { \alpha } \{ ( \mathbf { H } k + \mathbf { I } ) \cos \mathbf { J } x - 1 \} d x \\ & = \frac { \mathbf { K } } { \mathbf { L } } \{ \sqrt { k ( k + \mathbf { M } ) } - \alpha \} , \\ & S _ { 2 } = \frac { \mathbf { N } } { \mathbf { O } } \{ \sqrt { k ( k + \mathbf{P} ) } - \alpha \} + \frac { \pi } { \mathbf { Q } } . \end{aligned}$$
Hence, we obtain
$$S _ { 2 } - S _ { 1 } = \frac { \pi } { \mathbf { R } } ,$$
which shows that the value of $S _ { 2 } - S _ { 1 }$ is a constant independent of the value of $k$.