The value of \(\int_{0}^{\pi/2}\displaystyle\log_e\left(\frac{5+2\sin x}{5+2\cos x}\right)dx\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
Let I be the integral. By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx\) with \(a = \frac{\pi}{2}\), replacing x by \(\frac{\pi}{2} - x\) makes sin x and cos x interchange:
\(\displaystyle I = \int_{0}^{\pi/2} \log_e\left(\frac{5+2\cos x}{5+2\sin x}\right)dx\)
The fraction is now upside down, and \(\log\frac{1}{t} = -\log t\):
\(\displaystyle I = -\int_{0}^{\pi/2} \log_e\left(\frac{5+2\sin x}{5+2\cos x}\right)dx = -I\)
\(2I = 0,\quad I = 0\)