Question

If the function \( f(x) \) is defined as \( f(x)=\begin{vmatrix}\frac{1}{1+e^{\sin x}} & \sin x \\ \sqrt{1-x^{2}} & 2\end{vmatrix} \). Then the value of \( \frac{1}{\pi}\int_{-\pi/2}^{\pi/2} f(x)\,dx \) is equal to ________.

Hint:

Expand the determinant and use the property \( \int_{-a}^{a} g(x)\,dx=\int_{0}^{a}\big(g(x)+g(-x)\big)dx \).

Answer 1

Correct Answer: 1

Expanding the determinant: \( f(x)=\dfrac{2}{1+e^{\sin x}}-\sin x\sqrt{1-x^{2}} \).
Let \( I=\int_{-\pi/2}^{\pi/2} f(x)\,dx \).
The term \( \sin x\sqrt{1-x^{2}} \) is an odd function, so its integral over the symmetric interval is 0.
For the first term, \( \int_{-\pi/2}^{\pi/2}\dfrac{2}{1+e^{\sin x}}dx=\int_{0}^{\pi/2}\left(\dfrac{2}{1+e^{\sin x}}+\dfrac{2}{1+e^{-\sin x}}\right)dx \).
Since \( \dfrac{1}{1+e^{t}}+\dfrac{1}{1+e^{-t}}=1 \), this equals \( \int_{0}^{\pi/2}2\,dx=\pi \).
Hence \( I=\pi \) and \( \dfrac{1}{\pi}I=1 \).
Answer: 1