Đáp án:

• Gợi ý: \( \left( \sin^2{x} \right)' = 2 \sin{x} \cos{x} \)

•  \( I = \int \sin{x} \cos{x} \left( 1 - \sin^2{x} \right) e^{\sin^2{x}} \, dx \)

    \(= \frac{1}{2} \int u' f(u) \, du \quad \text{với } u = \sin^2{x} \)

• Đặt \( t = \sin^2{x} \Rightarrow dt = 2 \sin{x} \cos{x} \, dx \)

• \( I = \frac{1}{2} \int (1 - t) e^t \, dt\)

• Đặt  \( \begin{cases} u = 1 - t \\ dv = e^t \, dt \end{cases} \quad \Rightarrow \quad \begin{cases} du = -dt \\ v = e^t \end{cases} \)

• \( I = \frac{1}{2} \left[ (1 - t) e^t + \int e^t \, dt \right] \)

   \(= \frac{1}{2} \left[ (1 - t) e^t + e^t \right] + C \)

   \(= \frac{1}{2} \left[ (1 - \sin^2{x}) e^{\sin^2{x}} + e^{\sin^2{x}} \right] + C \)