Cho hàm số $\Large y = f(x)$ có $\Large f(0)= 0$ và $\Large f'(x) = \s

Cho hàm số $\Large  y = f(x)$ có $\Large  f(0)= 0$ và $\Large  f'(x) = \sin^{8}x - \cos^{8}x - 4\sin^{6}x, \forall x\in\mathbb{R}$

Tính $\Large  I = \int_0^{\pi}16f(x)dx.$

• A. A. $\Large  I = 160\pi$
• B. B. $\Large  I = -10\pi^{2}$
• C. C. $\Large  I = 16\pi^{2}$
• D. D. $\Large  I = 10\pi^{2}$

$\Large  f'(x) = \sin^{8}x - \cos^{8}x - 4\sin^{6}x = (\sin^{4}x - \cos^{4}x)(\sin^{4}x + \cos^{4}x) - 4\sin^{6}x$

$\Large   = (\sin^{2}x - \cos^{2}x)\left[(\sin^{2}x + \cos^{2}x)^{2} - 2\sin^{2}x\cos^{2}x\right] - 4\sin^{6}x$

$\Large  = (\sin^{2}x - \cos^{2}x)[1 - 2\sin ^{2}x\cos^{2}x] - 4\sin^{6}x$

$\Large   = (2\sin^{2}x -1)(2\sin^{4}x - 2\sin^{2}x +1) - 4\sin^{6}x$

$\Large  = -6\sin^{4}x + 4\sin^{2}x -1$

$\Large  f(x) = \dfrac{1}{2}\sin(2x) - \dfrac{3}{16}\sin(4x) - \dfrac{5}{4}x + C$

Mà $\Large  f(0) = 0 \Leftrightarrow C = 0$

Như vậy $\Large  \int_0^{\pi}16f(x)dx = \int_0^{\pi}16\left(\dfrac{1}{2}\sin (2x)-\dfrac{3}{16} \sin (4x) - \dfrac{5}{4}x\right)dx = -10\pi ^{2}$