Trong các khẳng định sau, khẳng định nào sai ?
\(\frac{\sqrt{10}+2}{3}\int\limits^{\frac{\pi}{4}}_0\frac{\sin2xdx}{\sqrt{\cos^2x+4\sin^2x}}=\frac{2}{3}\) \(\frac{\sqrt{10}+2}{3}-\int\limits^{\frac{\pi}{4}}_0\frac{\sin2xdx}{\sqrt{\cos^2x+4\sin^2x}}=\frac{4}{3}\) \(\left(\int\limits^{\frac{\pi}{4}}_0\frac{\sin2xdx}{\sqrt{\cos^2x+4\sin^2x}}\right)^2=1\) \(\int\limits^{\frac{\pi}{4}}_0\frac{3\sin2xdx}{\sqrt{\cos^2x+4\sin^2x}}+\int\limits^2_0dx=\sqrt{10}\) Hướng dẫn giải:Ta tính tích phân:
\(I=\int\limits^{\frac{\pi}{4}}_0\frac{\sin2xdx}{\sqrt{\cos^2x+4\sin^2x}}\)
Đặt \(t=\sqrt{\cos^2x+4\sin^2x}\) \(\Rightarrow t^2=\cos^2x+4\sin^2x\)
\(\Rightarrow2t\text{d}t=\left(-2\cos x.\sin x+8\sin x.\cos x\right)\text{d}x=6\sin x\cos x\text{d}x=3\sin2x\text{d}x\)
Đổi cận: \(x|^{\frac{\pi}{4}}_0\Rightarrow t|^{\sqrt{\frac{5}{2}}}_1\)
Vậy ta có:
\(I=\int\limits^{\sqrt{\frac{5}{2}}}_1\frac{2}{3}\frac{\text{td}t}{t}=\frac{2}{3}\int\limits^{\sqrt{\frac{5}{2}}}_1\text{d}t=\frac{2}{3}t|^{\sqrt{\frac{5}{2}}}_1\)
\(=\frac{2}{3}\left(\sqrt{\frac{5}{2}}-1\right)=\frac{\sqrt{10}-2}{3}\)
Do đó:
\(\frac{\sqrt{10}+2}{3}.I=\frac{\sqrt{10}+2}{3}\frac{\sqrt{10}-2}{3}=\frac{10-4}{9}=\frac{2}{3}\), A đúng.
\(\frac{\sqrt{10}+2}{3}-I=\frac{\sqrt{10}+2}{3}-\frac{\sqrt{10}-2}{3}=\frac{4}{3}\), B đúng
\(I^2=\left(\frac{\sqrt{10}-2}{3}\right)^2\ne1\), C sai
\(3I+\int\limits^2_0\text{d}x=3\frac{\sqrt{10}-2}{3}+x|^2_0=\sqrt{10}-2+2=\sqrt{10}\), D đúng.