Question

The value of \(\int_0^{\frac{\pi}{2}} log (\frac{5 + 4 sinx}{5 + 4 cosx})dx\) is:

• A. 0
• B. 1
• C. 2
• D. \(\frac{1}{2}\)

Answer 1

The Correct Option is A

To solve the integral \(\int_0^{\frac{\pi}{2}} \log \left(\frac{5 + 4 \sin x}{5 + 4 \cos x}\right)\,dx\), we use the property of definite integrals: \(\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx\).

Applying this property, let \( I = \int_0^{\frac{\pi}{2}} \log \left(\frac{5 + 4 \sin x}{5 + 4 \cos x}\right)\,dx \).

Then, by the symmetry property, \( I = \int_0^{\frac{\pi}{2}} \log \left(\frac{5 + 4 \sin (\frac{\pi}{2} - x)}{5 + 4 \cos (\frac{\pi}{2} - x)}\right)\,dx \).

Using trigonometric identities, \(\sin(\frac{\pi}{2} - x) = \cos x\) and \(\cos(\frac{\pi}{2} - x) = \sin x\), we transform the expression: \(\frac{5 + 4 \cos x}{5 + 4 \sin x}\).

Thus, the integral becomes: \( I = \int_0^{\frac{\pi}{2}} \log \left(\frac{5 + 4 \cos x}{5 + 4 \sin x}\right)\,dx \).

Adding the original and transformed integrals, we have:

\(2I = \int_0^{\frac{\pi}{2}} \left( \log \left(\frac{5 + 4 \sin x}{5 + 4 \cos x}\right) + \log \left(\frac{5 + 4 \cos x}{5 + 4 \sin x}\right)\right)\,dx \).

The expressions inside the logarithms simplify to:

\(\log \left(\frac{(5 + 4 \sin x)(5 + 4 \cos x)}{(5 + 4 \cos x)(5 + 4 \sin x)}\right) = \log (1) = 0\).

Therefore, \(2I = \int_0^{\frac{\pi}{2}} 0\,dx = 0 \).

Hence, \(I = 0\).

The value of the integral is 0.