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Câu hỏi: Xét tích phân $I=\int_{0}^{\dfrac{\pi }{3}}{\dfrac{\sin 2x}{1+\cos x}}dx$. Nếu đặt $t=\cos x$ thì tích phân $I$ trở thành
A. $I=-\int_{\dfrac{1}{2}}^{1}{\dfrac{2t}{1+t}}\text{d}t$.
B. $I=\int_{\dfrac{1}{2}}^{1}{\dfrac{2t}{1+t}}\text{d}t$.
C. $I=-\int_{0}^{\dfrac{\pi }{3}}{\dfrac{2t}{1+t}}\text{d}t$.
D. $I=\int_{0}^{\dfrac{\pi }{3}}{\dfrac{2t}{1+t}}\text{d}t$.
$I=\int_{0}^{\dfrac{\pi }{3}}{\dfrac{\sin 2x}{1+\cos x}}\text{d}x=\int_{0}^{\dfrac{\pi }{3}}{\dfrac{2\sin x.\cos x}{1+\cos x}}\text{d}x$
Đặt $t=\cos x\Rightarrow \text{d}t=-\sin x\text{d}x$
Đổi cận: $x=0\Rightarrow t=1;x=\dfrac{\pi }{3}\Rightarrow t=\dfrac{1}{2}$
Ta được: $I=-\int_{1}^{\dfrac{1}{2}}{\dfrac{2t}{1+t}\text{d}t}=\int_{\dfrac{1}{2}}^{1}{\dfrac{2t}{1+t}\text{d}t}$
A. $I=-\int_{\dfrac{1}{2}}^{1}{\dfrac{2t}{1+t}}\text{d}t$.
B. $I=\int_{\dfrac{1}{2}}^{1}{\dfrac{2t}{1+t}}\text{d}t$.
C. $I=-\int_{0}^{\dfrac{\pi }{3}}{\dfrac{2t}{1+t}}\text{d}t$.
D. $I=\int_{0}^{\dfrac{\pi }{3}}{\dfrac{2t}{1+t}}\text{d}t$.
$I=\int_{0}^{\dfrac{\pi }{3}}{\dfrac{\sin 2x}{1+\cos x}}\text{d}x=\int_{0}^{\dfrac{\pi }{3}}{\dfrac{2\sin x.\cos x}{1+\cos x}}\text{d}x$
Đặt $t=\cos x\Rightarrow \text{d}t=-\sin x\text{d}x$
Đổi cận: $x=0\Rightarrow t=1;x=\dfrac{\pi }{3}\Rightarrow t=\dfrac{1}{2}$
Ta được: $I=-\int_{1}^{\dfrac{1}{2}}{\dfrac{2t}{1+t}\text{d}t}=\int_{\dfrac{1}{2}}^{1}{\dfrac{2t}{1+t}\text{d}t}$
Đáp án B.