Chọn phương án B.

Đặt $u=2\sin x+1\Rightarrow\mathrm{d}u=2\cos x\mathrm{d}x$ hay $\cos x\mathrm{d}x=\dfrac{1}{2}\mathrm{d}u$.

• $x=0\Rightarrow u=1$
• $x=\dfrac{\pi}{2}\Rightarrow u=3$

Khi đó ta có $$\begin{aligned}
\displaystyle\int\limits_{0}^{\tfrac{\pi}{2}}f\left(2\sin x+1\right)\cos x\mathrm{\,d}x&=\dfrac{1}{2}\displaystyle\int\limits_{1}^{3}f(u)\mathrm{\,d}u=\dfrac{1}{2}\displaystyle\int\limits_{1}^{3}f(x)\mathrm{\,d}x\\
&=\dfrac{1}{2}\displaystyle\int\limits_{1}^{2}f(x)\mathrm{\,d}x+\dfrac{1}{2}\displaystyle\int\limits_{2}^{3}f(x)\mathrm{\,d}x\\
&=\dfrac{1}{2}\displaystyle\int\limits_{1}^{2}\left(x^2-2x+3\right)\mathrm{\,d}x+\dfrac{1}{2}\displaystyle\int\limits_{2}^{3}\left(x^2-1\right)\mathrm{\,d}x\\
&=\dfrac{1}{2}\left(\dfrac{x^3}{3}-x^2+3x\right)\bigg|_1^2+\dfrac{1}{2}\left(\dfrac{x^3}{3}-x\right)\bigg|_2^3\\
&=\dfrac{7}{6}+\dfrac{8}{3}=\dfrac{23}{6}.
\end{aligned}$$