Question

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx\]

 

Answer 1

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{2} \sin x\left| \sin x \right|dx\]

\[ = \int_{- \frac{\pi}{4}}^0 \sin x\left| \sin x \right|dx + \int_0^\frac{\pi}{2} \sin x\left| \sin x \right|dx\]

\[ = \int_{- \frac{\pi}{4}}^0 \sin x\left( - \sin x \right)dx + \int_0^\frac{\pi}{2} \sin x\sin xdx ...................\left( \left| \sin x \right| = \begin{cases}\sin x, & 0 \leq x \leq \frac{\pi}{2} \\ - \sin x, & - \frac{\pi}{4} \leq x \leq 0\end{cases} \right)\]

\[ = - \int_{- \frac{\pi}{4}}^0 \sin^2 xdx + \int_0^\frac{\pi}{2} \sin^2 xdx\]

\[= - \int_{- \frac{\pi}{4}}^0 \frac{1 - \cos2x}{2}dx + \int_0^\frac{\pi}{2} \frac{1 - \cos2x}{2}dx\]

\[ = - \frac{1}{2} \int_{- \frac{\pi}{4}}^0 dx + \frac{1}{2} \int_{- \frac{\pi}{4}}^0 \cos2xdx + \frac{1}{2} \int_0^\frac{\pi}{2} dx - \frac{1}{2} \int_0^\frac{\pi}{2} \cos2xdx\]

\[ = \left.- \frac{1}{2} \times x\right|_{- \frac{\pi}{4}}^0 +\left. \frac{1}{2} \times \frac{\sin2x}{2}\right|_{- \frac{\pi}{4}}^0 + \left.\frac{1}{2} \times x\right|_0^\frac{\pi}{2} - \left.\frac{1}{2} \times \frac{\sin2x}{2}\right|_0^\frac{\pi}{2} \]

\[ = - \frac{1}{2}\left( 0 + \frac{\pi}{4} \right) + \frac{1}{4}\left( 0 + \sin\frac{\pi}{2} \right) + \frac{1}{2} \times \left( \frac{\pi}{2} - 0 \right) - \frac{1}{4}\left( sin\pi - 0 \right)\]

\[ = - \frac{\pi}{8} + \frac{1}{4}\left( 0 + 1 \right) + \frac{\pi}{4} - \frac{1}{4}\left( 0 - 0 \right)\]

\[ = \frac{\pi}{8} + \frac{1}{4}\]