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 $k$.