Lời giải

Đặt \( t = \frac{\pi}{2} - x \Rightarrow dt = -dx \)

\( \begin{cases}
x = 0 \Rightarrow t = \frac{\pi}{2} \\
x = \frac{\pi}{2} \Rightarrow t = 0
\end{cases} \)

\( I = \int_{\frac{\pi}{2}}^{0} \frac{1 + \sin^n \left(\frac{\pi}{2} - t\right)}{2 + \sin^m \left(\frac{\pi}{2} - t\right)} (-dt) = \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos^n t}{2 + \cos^m t} \, dt = a \)

\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{B}} \)

\( \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx = \int_{0}^{\frac{\pi}{2}} f(\cos x) \, dx \)