Q.Choose the correct answer: The value of $\int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right)dx$ is (A) 2 (B) $\frac{3}{4}$ (C) 0 (D) $-2$

The key idea is Definite Integral Symmetry: for an integral from $0$ to $\pi/2$, substituting $x \to \pi/2 - x$ often reveals that the integrand is antisymmetric, giving a zero result.

Step 1: Let

Step 2: Use the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$. Here $a = \pi/2$, so substitute $x \to \pi/2 - x$:

$I = \int_{0}^{\pi/2} \log\left(\frac{4+3\sin(\pi/2 - x)}{4+3\cos(\pi/2 - x)}\right) dx = \int_{0}^{\pi/2} \log\left(\frac{4+3\cos x}{4+3\sin x}\right) dx.$ …

The problem asks for the value of a definite integral from $0$ to $\pi/2$ of a logarithm of a ratio. When you see an integral with limits like $0$ to $\pi/2$ and the integrand involves $\sin x$ and $\cos x$ in a symmetric way, your first instinct should be to check for symmetry. The key property here is the substitution $x \to \frac{\pi}{2} - x$, which swaps $\sin x$ and $\cos x$.

Let’s see why this works.

1. Set up the integral. Let

2. Apply the substitution $x \to \frac{\pi}{2} - x$. When $x = 0$, the new variable is $\pi/2$; when $x = \pi/2$, the new variable is $0$. The differential $dx$ becomes $-dx$, but flipping the limits gives a positive sign. So:

3. Simplify the trigonometric expressions. Recall: $\sin(\pi/2 - x) = \cos x$ and $\cos(\pi/2 - x) = \sin x$. Therefore:

4. Observe the relationship between the two expressions for $I$. The logarithm of a reciprocal is the negative of the logarithm: …

Method: Zero-by-antisymmetry using $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$

Use this whenever a definite integral over a symmetric-looking interval has an integrand that flips sign under the reflection $x\to a-x$ — the integral must then be $0$, with no antiderivative needed.

Steps

Step 1: Name the integral and apply the reflection property.

For any integrable $f$,

Set $I$ equal to the given integral and rewrite it with $x\to a-x$ (here $a=\tfrac{\pi}{2}$, which swaps $\sin x\leftrightarrow\cos x$ since $\sin(\tfrac{\pi}{2}-x)=\cos x$).

Step 2: Check whether the new integrand is the negative of the original. …

Common Mistakes

Mistake 1: Trying to evaluate the log integral directly.

Why it's wrong: the antiderivative of $\log\!\bigl(\tfrac{4+3\sin x}{4+3\cos x}\bigr)$ is not elementary, so a head-on attack stalls. Correct approach: use the symmetry property $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$ and detect the sign flip instead of integrating.

Mistake 2: Not realising $x\to\tfrac{\pi}{2}-x$ turns the integrand into its own negative. …