Evaluate the integral: \[ I = \int_{-\pi}^{\pi} \frac{x \sin^3 x}{4 - \cos^2 x} dx. \]
Consider: \[ f(x) = \frac{x \sin^3 x}{4 - \cos^2 x} \] Examine its parity by replacing \( x \to -x \): \[ f(-x) = \frac{-x \cdot \sin^3(-x)}{4 - \cos^2(-x)} = \frac{-x \cdot (-\sin x)^3}{4 - \cos^2 x} = \frac{-x \cdot (-\sin^3 x)}{4 - \cos^2 x} = \frac{x \sin^3 x}{4 - \cos^2 x} = -f(x) \] So \( f(x) \) is an **odd function**.
For any odd function \( f(x) \), the integral over symmetric limits about zero is: \[ \int_{-a}^{a} f(x) \, dx = 0 \] Hence: \[ I = \int_{-\pi}^{\pi} \frac{x \sin^3 x}{4 - \cos^2 x} \, dx = 0 \]
You might have seen a result like: \[ I = 2\pi \left(1 - \frac{3}{4} \log 3\right) \] But this applies to **different** symmetric integrals that involve **even functions** or shift-symmetric expressions. In our case, the integrand is **odd** due to the factor \( x \sin^3 x \), hence: \[ I = -I \Rightarrow I = 0 \] No computation is required beyond parity analysis.
\( \boxed{0} \)
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |
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