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COMEDK 2025 Afternoon Solved Paper

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Question : 149 of 180

Marks: +1, -0

\int\limits_0^{\frac{\pi}{2}} \log \left(\frac{5+4 \sin x}{5+4 \cos x}\right) d x=

0
2
$-$ 2
$\frac{3}{4}$

Solution:

Concept:

Use the substitution

x \to \frac{\pi}{2} - x

to exploit the symmetry of the integrand.

Explanation:

Let

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

.

Apply the substitution

x = \frac{\pi}{2} - t

. Then

dx = -dt

, and the limits reverse:

x=0 \to t=\pi/2

,

x=\pi/2 \to t=0

.

Thus

I = \int_{\pi/2}^{0} \log \left( \frac{5+4\sin(\frac{\pi}{2}-t)}{5+4\cos(\frac{\pi}{2}-t)} \right) (-dt) = \int_0^{\pi/2} \log \left( \frac{5+4\cos t}{5+4\sin t} \right) \, dt

.

Since

\log \frac{a}{b} = -\log \frac{b}{a}

, we have

I = \int_0^{\pi/2} \left[ -\log \left( \frac{5+4\sin t}{5+4\cos t} \right) \right] dt = -I

.

Therefore

2I = 0

, so

I = 0

.

Answer:

0

, which corresponds to option A.

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