Integral

por **Fernandobertolaccini** » Sex Jul 18, 2014 01:06

Se f'(x)= ${cos}^{2}x$ -sen2x

e f $(\frac{\pi}{4}) = \frac{1}{4}$ , encontre f(x)

Resp: $\frac{x}{2}+\frac{sen2x}{2}+\frac{cos2x}{2}-\frac{\pi}{8}$

Muito obrigadoo !!

por **DanielFerreira** » Sáb Jul 19, 2014 22:28

Temos que $f(x) = \int f'(x)$ , então:

$\\ f(x) = \int f'(x) \\\\ f(x) = \int \cos^2 x - \sin (2x) dx \\\\\\ f(x) = \frac{x}{2} + \frac{\sin (2x)}{4} + \frac{\cos (2x)}{2} + c \\\\\\ f\left(\frac{\pi}{4}\right) = \frac{\frac{\pi}{4}}{2} + \frac{\sin \left(2 \cdot \frac{\pi}{4} \right)}{4} + \frac{\cos \left(2 \cdot \frac{\pi}{4} \right)}{2} + c \\\\\\ \frac{1}{4} = \frac{\pi}{8} + \frac{\sin \left(\frac{\pi}{2} \right)}{4} + \frac{\cos \left(\frac{\pi}{2} \right)}{2} + c \\\\\\ \cancel{\frac{1}{4}} = \frac{\pi}{8} + \cancel{\frac{1}{4}} + \frac{0}{2} + c \\\\ c = - \frac{\pi}{8}$

Por fim, $\boxed{f(x) = \frac{x}{2} + \frac{\sin (2x)}{4} + \frac{\cos (2x)}{2} - \frac{\pi}{8}}$

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