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Calculeze

\( \displaystyle \left|\frac {\int_0^{\frac {\pi}{2}} (x\cos x + 1)e^{\sin x}\ dx}{\int_0^{\frac {\pi}{2}} (x\sin x - 1)e^{\cos x}\ dx}\right|. \)

\( I_1={\int_0^{\frac%20{\pi}{2}}%20(x\cos%20x%20+%201)e^{\sin%20x}\%20dx} \\
I_2={\int_0^{\frac%20{\pi}{2}}%20(x\sin%20x%20-%201)e^{\cos%20x}\%20dx} \)

\( I_1=\int_0^{\frac{\pi}{2}}((\frac{\pi}{2}-x)\sin x +1)e^{\cos x} dx \Rightarrow I_1+I_2= \int_0^{\frac{\pi}{2}} \frac{\pi}{2} e^{\cos x} \sin x dx=\frac{\pi}{2}(e-1) \).

Ar mai trebui inca o ecuatie sa facem un sistem si gata... Dar acuma nu imi iese.

Integrand prin parti

\( I_1=xe^{\sin x}|_0^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}e^{\sin x}dx+\int_0^{\frac{\pi}{2}}e^{\sin x}dx=e\frac{\pi}{2} \)

Analog pentru \( I_2. \)