Question

The value of $\frac{16 S _{1}}{\pi}$ is_____

Answer 1

Correct Answer: 2

\(S_1 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^2 x \cdot 1 \, dx\)

=\(\frac{1}{2} \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} (1 - \cos 2x) \, dx\)

= \(\frac{1}{2} \left( x - \frac{{\sin 2x}}{x} \right)_{\frac{\pi}{8}}^{\frac{3\pi}{8}}\)

= \(\frac{1}{2} \left( \frac{\pi}{4} - 0 \right)\)

= \(\frac{\pi}{8}\)

\(\frac{{16S_1}}{{\pi}} = 2\)

Answer 2

\(S_2 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^2 x \left|4x - \pi\right| \, dx\)

=\(\int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} 4 \sin^2 x \left|x - \frac{\pi}{4}\right| \, dx\)

Let \(x - \frac{\pi}{4} = t\)

\(=> dx = dt\)

\(S_2 = \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 4 \sin^2 \left(\frac{\pi}{4} + t\right) \left|t\right| \, dt\)

= \(\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 2(1 - \cos 2(\frac{\pi}{4} + t)) \left|t\right| \, dt\)

= \(\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} (2 + 2 \sin 2t) \left|t\right| \, dt\)

=\(2\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} |t| \, dt + 2\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} |t| \sin 2t \, dt\)

= \(4\int_{0}^{\frac{\pi}{8}} t \, dt + 0\)

\(S_2 = \left[2t^2\right]_{0}^{\frac{\pi}{8}}\)

= \(\frac{\pi^2}{32}\)

\(\frac{48S_2}{\pi^2} = \frac{3}{2} = 1.5\)