Google
Câu hỏi: Giả sử hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$, thỏa mãn $f\left( \sin x+1 \right)=\cos x$ với mọi $x\in \mathbb{R}$, khi đó tích phân $\int\limits_{1}^{\dfrac{3}{2}}{f\left( x \right)\mathbf{d}x}$ bằng
A. $\dfrac{\pi }{12}+\dfrac{\sqrt{3}}{4}$.
B. $\dfrac{-\pi }{6}+\dfrac{\sqrt{3}}{4}$.
C. $\dfrac{\pi }{12}-\dfrac{\sqrt{3}}{8}$.
D. $\dfrac{\pi }{12}+\dfrac{\sqrt{3}}{8}$.
A. $\dfrac{\pi }{12}+\dfrac{\sqrt{3}}{4}$.
B. $\dfrac{-\pi }{6}+\dfrac{\sqrt{3}}{4}$.
C. $\dfrac{\pi }{12}-\dfrac{\sqrt{3}}{8}$.
D. $\dfrac{\pi }{12}+\dfrac{\sqrt{3}}{8}$.
$I=\int\limits_{1}^{\dfrac{3}{2}}{f\left( x \right)\mathbf{d}x}$
Đặt $x=\sin t+1\Rightarrow \mathbf{d}x=\cos t\mathbf{d}t$.
Đổi cận: $x=1\Rightarrow t=0; x=\dfrac{3}{2}\Rightarrow t=\dfrac{\pi }{6}$
Khi đó $I=\int\limits_{0}^{\dfrac{\pi }{6}}{f\left( \sin t+1 \right)}.\cos tdt=\int\limits_{0}^{\dfrac{\pi }{6}}{\cos t.\cos tdt=\int\limits_{0}^{\dfrac{\pi }{6}}{{{\cos }^{2}}tdt}}$
$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{6}}{\left( \dfrac{1}{2}+\dfrac{1}{2}\cos 2t \right)dt=\left. \left( \dfrac{1}{2}t+\dfrac{1}{4}\sin 2t \right) \right|}_{0}^{\dfrac{\pi }{6}}=\dfrac{\pi }{12}+\dfrac{\sqrt{3}}{8}$.
Đặt $x=\sin t+1\Rightarrow \mathbf{d}x=\cos t\mathbf{d}t$.
Đổi cận: $x=1\Rightarrow t=0; x=\dfrac{3}{2}\Rightarrow t=\dfrac{\pi }{6}$
Khi đó $I=\int\limits_{0}^{\dfrac{\pi }{6}}{f\left( \sin t+1 \right)}.\cos tdt=\int\limits_{0}^{\dfrac{\pi }{6}}{\cos t.\cos tdt=\int\limits_{0}^{\dfrac{\pi }{6}}{{{\cos }^{2}}tdt}}$
$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{6}}{\left( \dfrac{1}{2}+\dfrac{1}{2}\cos 2t \right)dt=\left. \left( \dfrac{1}{2}t+\dfrac{1}{4}\sin 2t \right) \right|}_{0}^{\dfrac{\pi }{6}}=\dfrac{\pi }{12}+\dfrac{\sqrt{3}}{8}$.
Đáp án D.