$\int\limits^\pi_0 {\sqrt[]{1-sin4x}} \, dx$ = a $\sqrt[]{b}$ Tìm a, b Giúp em bài này với ạ

Đáp án:

$a=2, b=2.$

Giải thích các bước giải:

$I=\displaystyle \int\limits^\pi_0 \sqrt{1-\sin 4x} \, dx\\ u=2x, du=2dx\\ \begin{array}{|c|c|} \hline x&0&\pi\\\hline u&0&2\pi \\\hline\end{array}\\ I=\dfrac{1}{2}\displaystyle \int\limits^{2\pi}_0 \sqrt{1-\sin 2u} \, du\\ =\dfrac{1}{2}\displaystyle \int\limits^{2\pi}_0 \sqrt{1-\sin 2x} \, dx\\ =\dfrac{1}{2}\displaystyle \int\limits^{2\pi}_0 \sqrt{\sin^2x-2\sin x\cos x+\cos^2x} \, dx\\ =\dfrac{1}{2}\displaystyle \int\limits^{2\pi}_0 \sqrt{(\sin x-\cos x)^2} \, dx\\ =\dfrac{1}{2}\displaystyle \int\limits^{2\pi}_0 \sqrt{2\sin^2\left(x-\dfrac{\pi}{4}\right)^2} \, dx\\ =\dfrac{\sqrt{2}}{2} \displaystyle \int\limits^{2\pi}_0 \left|\sin\left(x-\dfrac{\pi}{4}\right)\right| \, dx\\ =\dfrac{\sqrt{2}}{2} \displaystyle \int\limits^{\tfrac{7\pi}{4}}_{-\tfrac{\pi}{4}} \left|\sin x \right| \, dx\\ =\dfrac{\sqrt{2}}{2}\left(- \displaystyle \int\limits^{0}_{-\tfrac{\pi}{4}} \sin x \, dx+ \displaystyle \int\limits^{\pi}_{0} \sin x\, dx -\displaystyle \int\limits^{\tfrac{7\pi}{4}}_{\pi}\sin x \, dx\right)\\ =\dfrac{\sqrt{2}}{2}.4\\ =2\sqrt{2}\\ \Rightarrow a=2, b=2.$