Question

The absolute value of $\frac{\underset{0}{\overset{\pi /2}{\int }}(x\cos x+1)e^{\sin x}dx}{\underset{0}{\overset{\pi /2}{\int }}(x\sin x-1)e^{\cos x}dx}$ is equal to

• A. $e$
• B. $\pi e$
• C. $\frac{e}{2}$
• D. $\frac{\pi }{e}$

Answer 1

The correct option is A $e$

$I_{Nr}=\underset{0}{\overset{\pi /2}{\int }}\underset{\text{I}}{\underset{}{x}}\cdot \underset{\text{II}}{\underset{}{\cos x{e}^{\sin x}}}dx+\underset{0}{\overset{\pi /2}{\int }}{e}^{\sin x}dx$
$={\left[x{e}^{\sin x}\right]}_0^{\pi /2}-\underset{0}{\overset{\pi /2}{\int }}{e}^{\sin x}dx+\underset{0}{\overset{\pi /2}{\int }}{e}^{\sin x}dx$
$\Rightarrow I_{Nr}=\frac{e\pi }{2}\cdots \left(1\right)$

Again,
$I_{Dr}=\underset{0}{\overset{\pi /2}{\int }}\underset{\text{I}}{\underset{}{x}}\cdot \underset{\text{II}}{\underset{}{\sin x{e}^{\cos x}}}dx-\underset{0}{\overset{\pi /2}{\int }}{e}^{\cos x}dx$
$={[-x{e}^{\cos x}]}_0^{\pi /2}+\underset{0}{\overset{\pi /2}{\int }}{e}^{\cos x}dx-\underset{0}{\overset{\pi /2}{\int }}{e}^{\cos x}dx$
$\Rightarrow I_{Dr}=-\frac{\pi }{2}\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$\left|\frac{I_{Nr}}{I_{Dr}}\right|=\frac{e\pi \cdot 2}{2\pi }=e$