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Câu hỏi: Xét tích phân $I=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{\sin 2x}{1+\cos x}dx}$. Nếu đặt $t=\cos x$ thì tích phân $I$ trở thành
A. $I=-\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{2t}{1+t}}dt$.
B. $I=-\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{2t}{1+t}}dt$.
C. $I=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{2t}{1+t}}dt$.
D. $I=\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{2t}{1+t}}dt$.
A. $I=-\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{2t}{1+t}}dt$.
B. $I=-\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{2t}{1+t}}dt$.
C. $I=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{2t}{1+t}}dt$.
D. $I=\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{2t}{1+t}}dt$.
Đặt $t=\cos x\Rightarrow dt=-\sin xdx\Rightarrow -dt=\sin x dx$.
Đổi cận:
$\begin{aligned}
& x=\dfrac{\pi }{3}\Rightarrow t=\dfrac{1}{2} \\
& x=\dfrac{\pi }{2}\Rightarrow t=0 \\
\end{aligned}$
$I=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{\sin 2x}{1+\cos x}dx}=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{2\sin x.\cos x}{1+\cos x}dx}=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{2\cos x}{1+\cos x}\sin xdx}=\int\limits_{\dfrac{1}{2}}^{0}{\dfrac{2t}{1+t}}\left( -dt \right)=-\int\limits_{\dfrac{1}{2}}^{0}{\dfrac{2t}{1+t}}dt=\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{2t}{1+t}}dt$.
Đổi cận:
$\begin{aligned}
& x=\dfrac{\pi }{3}\Rightarrow t=\dfrac{1}{2} \\
& x=\dfrac{\pi }{2}\Rightarrow t=0 \\
\end{aligned}$
$I=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{\sin 2x}{1+\cos x}dx}=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{2\sin x.\cos x}{1+\cos x}dx}=\int\limits_{\dfrac{\pi }{3}}^{\dfrac{\pi }{2}}{\dfrac{2\cos x}{1+\cos x}\sin xdx}=\int\limits_{\dfrac{1}{2}}^{0}{\dfrac{2t}{1+t}}\left( -dt \right)=-\int\limits_{\dfrac{1}{2}}^{0}{\dfrac{2t}{1+t}}dt=\int\limits_{0}^{\dfrac{1}{2}}{\dfrac{2t}{1+t}}dt$.
Đáp án D.