Tích phân \(\int\limits^{\frac{\pi}{4}}_0\frac{\sin\left(x-\frac{\pi}{4}\right)}{\sin2x+2\left(1+\sin x+\cos x\right)}\text{d}x\) bằng
\(\frac{4+3\sqrt{2}}{4}\). \(\frac{-4+3\sqrt{2}}{4}\). \(\frac{4-3\sqrt{2}}{4}\). \(\frac{-4-3\sqrt{2}}{4}\) Hướng dẫn giải:Ta có:
\(\sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\left(\sin x-\cos x\right)\)
\(\sin2x+2\left(1+\sin x+\cos x\right)=2\sin x\cos x+2+\sin x+\cos x\)
\(=\sin^2x+\cos^2x+1+2\sin x\cos x+2\sin x+2\cos x=\left(1+\sin x+\cos x\right)^2\)
\(I=\int\limits^{\frac{\pi}{4}}_0\frac{\sin\left(x-\frac{\pi}{4}\right)}{\sin2x+2\left(1+\sin x+\cos x\right)}\text{d}x=\int\limits^{\frac{\pi}{4}}_0\frac{\frac{\sqrt{2}}{2}\left(\sin x-\cos x\right)}{\left(1+\sin x+\cos x\right)^2}\text{d}x\)
Ta đặt \(t=1+\sin x+\cos x\)
⇒ \(\text{d}t=\left(\cos x-\sin x\right)\text{d}x=-\left(\sin x-\cos x\right)\text{d}x\)\(I=\int\limits^{1+\sqrt{2}}_2\frac{\frac{\sqrt{2}}{2}\left(-\text{d}t\right)}{t^2}\)
\(=\frac{\sqrt{2}}{2}\frac{1}{t}|^{1+\sqrt{2}}_2\)
\(=\frac{\sqrt{2}}{2}\left[\frac{1}{1+\sqrt{2}}-\frac{1}{2}\right]\)\(=\frac{4-3\sqrt{2}}{4}\).