Question

If $I_1={\int }_0^{\pi }x\sin x{e}^{\cos 4x}dx\&I_2={\int }_0^{\frac{\pi }{2}}\cos x{e}^{\cos 4x}dx$, then the value of $[\frac{I_1}{I_2}]$ is (where [.] denotes the greatest integer function)

Answer 1

$I_1={\int }_0^{\pi }x\sin x{e}^{\cos 4x}dx\cdots$ (i)
$I_1={\int }_0^{\pi }(\pi -x)\sin x{e}^{\cos 4x}dx\cdots$ (ii)
On adding (i) and (ii), we get-
$2I_1=\pi {\int }_0^{\pi }\sin x.e^{\cos 4x}dx2I_1=2\pi {\int }_0^{\frac{\pi }{2}}\sin x{e}^{\cos 4x}dx{I}_1=\pi {\int }_0^{\frac{\pi }{2}}\sin x.e^{\cos 4x}dx{I}_1=\pi {\int }_0^{\frac{\pi }{2}}\cos x.e^{\cos 4x}dx\frac{I_1}{I_2}=\pi \Rightarrow [\frac{I_1}{I_2}]=3$