Tính tích phân trên đoạn [0;2pi] của căn (1+sin x) dx

Đáp án:

$4\sqrt2$

Giải thích các bước giải:

$\quad I =  \displaystyle\int\limits_0^{2\pi}\sqrt{1 + \sin x}dx$

$\to I = \displaystyle\int\limits_0^{2\pi}\sqrt{1 + \cos\left(\dfrac{\pi}{2} - x\right)}dx$

$\to I = \displaystyle\int\limits_0^{2\pi}\sqrt2.\sqrt{\cos^2\left(\dfrac{\pi}{4}-\dfrac x2\right)}dx$

$\to I = \sqrt2\displaystyle\int\limits_0^{2\pi}\left|\cos\left(\dfrac{\pi}{4}-\dfrac x2\right)\right|dx$

Đặt $u = \dfrac{\pi}{4}-\dfrac x2$

$\to du =-\dfrac12dx$

Ta được:

$\quad I = -2\sqrt2\displaystyle\int\limits_{\tfrac{\pi}{4}}^{-\tfrac{3\pi}{4}}\left|\cos u\right|du$

$\to I = 2\sqrt2\displaystyle\int\limits^{\tfrac{\pi}{4}}_{-\tfrac{3\pi}{4}}\left|\cos u\right|du$

$\to I = 2\sqrt2\displaystyle\int\limits^{\tfrac{\pi}{2}}_{-\tfrac{\pi}{2}}\left|\cos u\right|du$

$\to I = 2\sqrt2\displaystyle\int\limits^{\tfrac{\pi}{2}}_{-\tfrac{\pi}{2}}\cos udu$

$\to I = 2\sqrt2\sin u\Bigg|^{\tfrac{\pi}{2}}_{-\tfrac{\pi}{2}}$

$\to I = 4\sqrt2$