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Câu hỏi: Xét $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin 2x}{\sqrt{1+\cos x}} }\text{d}x$. Nếu đặt $t=\sqrt{1+\cos x}$, ta được :
A. $I=-4\int\limits_{1}^{\sqrt{2}}{\left({{t}^{2}}-1 \right) \text{d}}t$.
B. $I=4\int\limits_{1}^{\sqrt{2}}{\left({{t}^{2}}-1 \right) \text{d}t}$
C. $I=-\int\limits_{1}^{\sqrt{2}}{\dfrac{4{{t}^{3}}-4t}{t} \text{d}}t$.
D. $I=\int\limits_{\sqrt{2}}^{1}{\dfrac{-4{{t}^{3}}+4t}{t} \text{d}}t$.
A. $I=-4\int\limits_{1}^{\sqrt{2}}{\left({{t}^{2}}-1 \right) \text{d}}t$.
B. $I=4\int\limits_{1}^{\sqrt{2}}{\left({{t}^{2}}-1 \right) \text{d}t}$
C. $I=-\int\limits_{1}^{\sqrt{2}}{\dfrac{4{{t}^{3}}-4t}{t} \text{d}}t$.
D. $I=\int\limits_{\sqrt{2}}^{1}{\dfrac{-4{{t}^{3}}+4t}{t} \text{d}}t$.
Vìđặt $t=\sqrt{1+\cos x}$ $\Rightarrow {{t}^{2}}=1+\cos x\Rightarrow 2t\text{d}t=-\sin x \text{d}x$.
Đổi cận : $x=0\Rightarrow t=\sqrt{2}$
$x=\dfrac{\pi }{2}\Rightarrow t=1$
Khi đó :
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin 2x}{\sqrt{1+\cos x}}} \text{d}x=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\sin x.\cos x}{\sqrt{1+\cos x}} }\text{d}x =\int\limits_{1}^{\sqrt{2}}{\dfrac{2\left( {{t}^{2}}-1 \right).2t}{t}\text{d}t}\text{=}4\int\limits_{1}^{\sqrt{2}}{\left( {{t}^{2}}-1 \right) \text{d}t}$.
Chọn $B$.
Đổi cận : $x=0\Rightarrow t=\sqrt{2}$
$x=\dfrac{\pi }{2}\Rightarrow t=1$
Khi đó :
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin 2x}{\sqrt{1+\cos x}}} \text{d}x=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{2\sin x.\cos x}{\sqrt{1+\cos x}} }\text{d}x =\int\limits_{1}^{\sqrt{2}}{\dfrac{2\left( {{t}^{2}}-1 \right).2t}{t}\text{d}t}\text{=}4\int\limits_{1}^{\sqrt{2}}{\left( {{t}^{2}}-1 \right) \text{d}t}$.
Chọn $B$.
Đáp án B.