Question

If $I_m=\underset{0}{\overset{m\pi }{\int }}x\left|\sin x\right|e^{\cos 4x}dx$ and $J_n=\underset{0}{\overset{n\pi }{\int }}\left|\cos x\right|e^{\cos 4x}dx,$ where $m,n\in Z,$ then
(Here, $\left[k\right]$ denotes the greatest integer less than or equal to $k$.)

• A. $\left[\frac{I_1}{J_1}\right]=1$
• B. $\left[\frac{I_2}{J_2}\right]=3$
• C. $\left[\frac{I_2}{J_1}\right]=0$
• D. $\left[\frac{I_3}{J_3}\right]=4$

Answer 1

Answer: (A), (B), (D)

$\left[\frac{I_3}{J_3}\right]=4$

$I_m=\underset{0}{\overset{m\pi }{\int }}x\left|\sin x\right|e^{\cos 4x}dx$
$x\to m\pi -x$, we get
$I_m=\underset{0}{\overset{m\pi }{\int }}(m\pi -x)\left|\sin x\right|e^{\cos 4x}dx$
Adding both, we get
$2I_m=m\pi \underset{0}{\overset{m\pi }{\int }}\left|\sin x\right|e^{\cos 4x}dx\Rightarrow 2I_m=m^2\pi \underset{0}{\overset{\pi }{\int }}\left|\sin x\right|e^{\cos 4x}dx\Rightarrow 2I_m=2m^2\pi \underset{0}{\overset{\pi /2}{\int }}\sin x{e}^{\cos 4x}dx\therefore I_m=m^2\pi \underset{0}{\overset{\pi /2}{\int }}\sin x{e}^{\cos 4x}dx$

Now, $J_n=\underset{0}{\overset{n\pi }{\int }}\left|\cos x\right|e^{\cos 4x}dx$
$\Rightarrow J_n=n\underset{0}{\overset{\pi }{\int }}\left|\cos x\right|e^{\cos 4x}dx\Rightarrow J_n=2n\underset{0}{\overset{\pi /2}{\int }}\cos x{e}^{\cos 4x}dx\therefore J_n=2n\underset{0}{\overset{\pi /2}{\int }}\sin x{e}^{\cos 4x}dx$

Therefore, $\frac{I_m}{J_n}=\frac{m^2\pi }{2n}$
$\frac{I_1}{J_1}=\frac{\pi }{2}\Rightarrow \left[\frac{I_1}{J_1}\right]=1\frac{I_2}{J_2}=\frac{4\pi }{4}=\pi \Rightarrow \left[\frac{I_2}{J_2}\right]=3\frac{I_2}{J_1}=\frac{4\pi }{2}=2\pi \Rightarrow \left[\frac{I_2}{J_1}\right]=6\frac{I_3}{J_3}=\frac{9\pi }{6}=\frac{3\pi }{2}\Rightarrow \left[\frac{I_3}{J_3}\right]=4$