Xét tích phân $\Large I=\int\limits_0^{\dfrac{\pi}{2}}\dfrac{\sin{2x}}

Xét tích phân $\Large I=\int\limits_0^{\dfrac{\pi}{2}}\dfrac{\sin{2x}}{\sqrt{1+\cos{x}}}\mathrm{d}x.$ Nếu đặt $\Large t=\sqrt{1+cosx},$ ta được

• A.

  A. $\Large I=-4\int\limits_1^{\sqrt{2}}(t^2-1)\mathrm{d}t.$

• B.

  B. $\Large I=4\int\limits_1^{\sqrt{2}}(t^2-1)\mathrm{d}t.$

• C.

  C. $\Large I=\int\limits_{\sqrt{2}}^1\dfrac{4t^3-4t}{t}\mathrm{d}t.$

• D.

  D. $\Large I=\int\limits_{\sqrt{2}}^1\dfrac{-4t^3+4t}{t}\mathrm{d}t.$

Chọn B

Đặt $\Large t=\sqrt{1+\cos{x}}$ $\Large \Rightarrow \cos{x}=t^2-1$ $\Large \Rightarrow \sin{x}.\mathrm{d}x=-2t.\mathrm{d}t.$

Đổi cận: $\Large x=0 \Rightarrow t=\sqrt{2}$; $\Large x=\dfrac{\pi}{2} \Rightarrow t=1.$

Khi đó ta có: $\Large I=\int\limits_0^{\dfrac{\pi}{2}}\dfrac{2\sin{x}\cos{x}}{\sqrt{1+\cos{x}}}\mathrm{d}x$ $\Large =\int\limits_{\sqrt{2}}^1\dfrac{2(t^2-1)(-2t\mathrm{d}t)}{t}$ $\Large =-4\int\limits_{\sqrt{2}}^1(t^2-1)\mathrm{d}t$ $\Large =4\int\limits_1^{\sqrt{2}}(t^2-1)\mathrm{d}t.$