Question

\( \int_0^{\pi} x \left( \sin^2(\sin x) + \cos^2(\cos x) \right) dx = \)

• A. \( \pi^2 \)
• B. \( \frac{\pi^2}{2} \)
• C. \( 2\pi \)
• D. \( \frac{\pi}{4} \)

Hint:

Use symmetry \( f(x) + f(\pi - x) \) tricks when dealing with definite integrals over \( [0, \pi] \).

Answer 1

The Correct Option is B

Observe that: \[ \sin^2(\sin x) + \cos^2(\cos x) \leq 1 + 1 = 2,\ \text{but not a standard simplification} \] However, the key is symmetry. Let \( f(x) = x (\sin^2(\sin x) + \cos^2(\cos x)) \).
Check \( f(\pi - x) \): \[ f(\pi - x) = (\pi - x) (\sin^2(\sin (\pi - x)) + \cos^2(\cos (\pi - x))) = (\pi - x)(\sin^2(\sin x) + \cos^2(\cos x)) \] So, \( f(\pi - x) + f(x) = \pi (\sin^2(\sin x) + \cos^2(\cos x)) \).
Thus, \[ \int_0^{\pi} f(x) dx = \frac{1}{2} \int_0^{\pi} \left[ f(x) + f(\pi - x) \right] dx = \frac{1}{2} \int_0^{\pi} \pi (\sin^2(\sin x) + \cos^2(\cos x)) dx \] \[ = \frac{\pi}{2} \int_0^{\pi} (\sin^2(\sin x) + \cos^2(\cos x)) dx \] Since average value of these periodic expressions over \( [0, \pi] \approx 1 \), so integral \( \approx \pi \), thus: \[ \text{Answer} = \frac{\pi^2}{2} \]