MathematicsEasy210×since 2002Q3440Let I=∫π/4π/3(8sinx−sin2xx)dxI=\int_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d xI=∫π/4π/3(x8sinx−sin2x)dx. ThenAπ2<I<3π4{\pi \over 2} < I < {{3\pi } \over 4}2π<I<43πBπ5<I<5π12{\pi \over 5} < I < {{5\pi } \over {12}}5π<I<125πC5π12<I<23π{{5\pi } \over {12}} < I < {{\sqrt 2 } \over 3}\pi125π<I<32πD3π4<I<π{{3\pi } \over 4} < I < \pi43π<I<πCheck answerSkip