Hướng dẫn giải (chi tiết)

Chọn D. 25.

$\displaystyle\int\limits_{0}^{\frac{\pi}{2}}\sin^2\dfrac{x}{4}\cos^2\dfrac{x}{4}dx$

$=\dfrac{1}{4}\displaystyle\int\limits_{0}^{\frac{\pi}{2}}\sin^2\dfrac{x}{2}dx$

$=\dfrac{1}{4}\displaystyle\int\limits_{0}^{\frac{\pi}{2}}\dfrac{1-\cos x}{2}dx$

$=\dfrac{1}{8}(x-\sin x)\;\bigg|_0^{\frac{\pi}{2}}$

$=\dfrac{\pi}{16}-\dfrac{1}{8}$

Khi đó a = 1, b = 8, c = 16 nên P = a + b + c = 1 + 8 + 16 = 25.

Hướng dẫn giải (chi tiết)

Chọn B. $4\sqrt{2}$.

Với x ∈ [0 ; $\pi$] thì sin x $\ge$ 0 nên |sin x| = sin x.

Với x ∈ [$\pi$ ; 2$\pi$] thì sin x $\le$ 0 nên |sin x| = –sin x.

Mà $\sqrt{1-\cos2x}=\sqrt{2\sin^2x}=\sqrt{2}|\sin x|$ nên 

$I=\displaystyle\int\limits_{0}^{2\pi}\sqrt{1-\cos2x}dx$

$=\sqrt{2}\displaystyle\int\limits_{0}^{2\pi}|\sin x|dx$

$=\sqrt{2}\bigg(\displaystyle\int\limits_{0}^{\pi}\sin xdx-\displaystyle\int\limits_{\pi}^{2\pi}\sin xdx\bigg)$

$=\sqrt{2}\bigg(-\cos x\bigg|_0^\pi+\cos x\bigg|_\pi^{2\pi}\bigg)$

$=4\sqrt{2}$.