$\int_{-\pi}^{\pi} (\cos 2x \cdot \cos^2 2x \cdot \cos^3 x \cdot \cos^2 4x \cdot \cos^2 5x)\, dx$
$\int_{-1}^{1} \ln(x + \sqrt{x^2+1})\, dx$
$\int_0^1 \tan^{-1}\left(\dfrac{2x-1}{1+x+x^2}\right) dx$
$\int_0^{\pi/2} \ln(\tan x)\, dx$
Step-by-Step Solution
Key Concept: The key idea is to evaluate definite integrals for vanishing conditions by leveraging their fundamental properties: checking for odd functions over symmetric intervals, applying King's Rule (the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$), or simplifying the integrand using appropriate trigonometric and inverse trigonometric identities.
Step 1: Check option (a): $\int_{-\pi}^{\pi} \cos 2x \cdot \cos^2 2x \cdot \cos^3 x \cdot \cos^2 4x \cdot \cos^2 5x\, dx$. The integrand contains $\cos^3 x$ (odd power of cosine) multiplied by even functions. The product $\cos 2x \cdot \cos^2 2x \cdot \cos^2 4x \cdot \cos^2 5x$ is even, and $\cos^3 x$ is even. So the full integrand is even and the integral does not necessarily vanish.
Step 2: Check option (b): $\int_{-1}^{1} \ln(x + \sqrt{x^2+1})\, dx$. Let $h(x) = \ln(x+\sqrt{x^2+1})$. Note $h(-x) = \ln(-x+\sqrt{x^2+1}) = -\ln(x+\sqrt{x^2+1}) = -h(x)$. So $h(x)$ is an odd function. The integral of an odd function over $[-1,1]$ is 0. ✓
Step 3: Check option (c): $\int_0^1 \tan^{-1}\left(\dfrac{2x-1}{1+x+x^2}\right) dx$. Note $\dfrac{2x-1}{1+x+x^2} = \dfrac{x-(1-x)}{1+x(1-x)} = \tan^{-1}(x) - \tan^{-1}(1-x)$ (using $\tan(A-B)$ formula). So the integral becomes $\int_0^1 [\tan^{-1}(x) - \tan^{-1}(1-x)]\, dx$. By substitution $x \to 1-x$, $\int_0^1 \tan^{-1}(x)\,dx = \int_0^1 \tan^{-1}(1-x)\,dx$. Hence the integral = 0. ✓
Step 4: Check option (d): $\int_0^{\pi/2} \ln(\tan x)\, dx$. Using the property $\int_0^{\pi/2} f(x)\,dx = \int_0^{\pi/2} f(\pi/2 - x)\,dx$:
$$I = \int_0^{\pi/2} \ln(\tan(\pi/2 - x))\,dx = \int_0^{\pi/2} \ln(\cot x)\,dx = -\int_0^{\pi/2} \ln(\tan x)\,dx = -I$$
So $2I = 0 \Rightarrow I = 0$. ✓
Step 5: Options (b), (c), and (d) all vanish.
Correct Answer: 2, 3, 4