Lời giải

\( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x}{4 - \sin^2 x} \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{4 - \sin^2 x} \, dx \)

\(= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{4 - \sin^2 x} \, dx = \int_{-1}^1 \frac{1}{4 - t^2} \, dt = \int_{-1}^1 \frac{1}{(2 - t)(2 + t)} \, dt\)

\(= \frac{1}{4}\int_{-1}^1\left ( \frac{1}{2 - t} + \frac{1}{2 + t} \, dt \right) = \frac{1}{4} \left[\ln |2 + t| - \ln |2 - t| \right]\bigg|_{-1}^1 \)

\( = \frac{1}{4} \ln 3 + \frac{1}{4} \ln 3 = \frac{1}{2} \ln 3\)

Bấm: \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x + \cos x}{4 - \sin^2 x} \, dx = 0.549306 = \frac{1}{2} \ln 3\)