Question

The absolute value of $\frac{{\int }_0^{\pi /2}\left(x\cos x+1\right)e^{\sin x}dx}{{\int }_0^{\pi /2}\left(x\sin x-1\right)e^{\cos x}dx}$ is equal to

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

Answer 1

The correct option is A $e$

$\begin{array}{c}Wehave\\ I=\frac{{\int }_0^{\frac{\pi }{2}}{e}^{\sin x}\left(x\cos x+1\right)dx}{{\int }_0^{\frac{\pi }{2}}{e}^{\cos x}\left(x\sin x-1\right)dx}\\ =\frac{{\left[x{e}^{\sin x}\right]}_0^{\frac{\pi }{2}}}{{\int }_0^{\frac{\pi }{2}}{e}^{\cos x}\left(1-x\sin x\right)dx}=\frac{\frac{\pi }{2}\times e}{{\left[e^{\cos x}x\right]}_{\frac{\pi }{2}}^0}\\ =\frac{\frac{\pi }{2}e}{0-\frac{\pi }{2}}=-e\\ Hence,absolutevalue=e\\ Hence,optionAisthecorrectanswer.\end{array}$