Question

Evaluate: \[ I = \int_{0}^{\pi/2} \frac{ x \sin x \cos x }{ \sin^4 x + \cos^4 x } \, dx. \]

Hint:

For definite integrals with $\sin x$ and $\cos x$, try $f(x) + f(\frac{\pi}{2} - x)$ trick.

Answer 1

Step 1: Use the substitution for definite integrals: \[ I = \int_{0}^{\pi/2} f(x) dx. \text{Try } x \mapsto \frac{\pi}{2} - x. \] Let’s check: \[ f\Big( \frac{\pi}{2} - x \Big) = \frac{ \Big( \frac{\pi}{2} - x \Big) \sin(\frac{\pi}{2} - x) \cos(\frac{\pi}{2} - x) } { \sin^4(\frac{\pi}{2} - x) + \cos^4(\frac{\pi}{2} - x) }. \] But: \[ \sin(\frac{\pi}{2} - x) = \cos x, \cos(\frac{\pi}{2} - x) = \sin x. \] So, \[ f\Big( \frac{\pi}{2} - x \Big) = \frac{ [\frac{\pi}{2} - x] \sin x \cos x }{ \cos^4 x + \sin^4 x }. \] So: \[ I = \int_{0}^{\pi/2} f(x) dx = \int_{0}^{\pi/2} f\Big( \frac{\pi}{2} - x \Big) dx. \] Add: \[ 2I = \int_{0}^{\pi/2} [f(x) + f(\frac{\pi}{2} - x)] dx. \] So, \[ 2I = \int_{0}^{\pi/2} \Bigg[ \frac{ x \sin x \cos x + (\frac{\pi}{2} - x)\sin x \cos x }{ \sin^4 x + \cos^4 x } \Bigg] dx. \] Combine numerator: \[ 2I = \int_{0}^{\pi/2} \frac{ \frac{\pi}{2} \sin x \cos x }{ \sin^4 x + \cos^4 x } dx. \] So, \[ 2I = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{ \sin x \cos x }{ \sin^4 x + \cos^4 x } dx. \] Step 2: Substitute $\sin^2 x = t$. \[ \text{Let } t = \sin^2 x, dt = 2 \sin x \cos x\, dx. \] So, \[ \sin x \cos x\, dx = \frac{1}{2} dt. \text{When } x = 0, t = 0; x = \frac{\pi}{2} \implies t = 1. \] So, \[ 2I = \frac{\pi}{2} \times \frac{1}{2} \int_{0}^{1} \frac{dt}{t^2 + (1 - t)^2}. \] Step 3: Expand denominator: \[ t^2 + (1 - t)^2 = t^2 + 1 - 2t + t^2 = 2t^2 - 2t + 1. \] So, \[ 2I = \frac{\pi}{4} \int_{0}^{1} \frac{dt}{2t^2 - 2t + 1} = \frac{\pi}{4} \int_{0}^{1} \frac{dt}{2(t^2 - t + \frac{1}{2})}. \] Complete square: \[ t^2 - t + \frac{1}{2} = t^2 - t + \frac{1}{4} + \frac{1}{4} = (t - \frac{1}{2})^2 + \frac{1}{4}. \] So, \[ 2I = \frac{\pi}{8} \int_{0}^{1} \frac{dt}{ (t - \frac{1}{2})^2 + \frac{1}{4} }. \] Step 4: Use formula: $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a}$. Put $u = t - \frac{1}{2}$: \[ 2I = \frac{\pi}{8} \int_{-1/2}^{1/2} \frac{du}{u^2 + (1/2)^2} = \frac{\pi}{8} \times \frac{1}{\frac{1}{2}} \Big[ \tan^{-1} (2u) \Big]_{-1/2}^{1/2}. \] \[ = \frac{\pi}{4} [\tan^{-1} 1 - \tan^{-1}(-1)] = \frac{\pi}{4} [\frac{\pi}{4} + \frac{\pi}{4}] = \frac{\pi}{4} \times \frac{\pi}{2} = \frac{\pi^2}{8}. \] Therefore, \[ I = \frac{\pi^2}{16}. \] Final Answer: $\boxed{\frac{\pi^2}{16}}$