# Evaluate: integration[sin(x^0)dx]?

Apr 7, 2018

$\left(\sin 1\right) \cdot x + C$

or

$- \frac{\pi}{180} \cos \left(\frac{180}{\pi} x\right) + C$

#### Explanation:

if You mean $x$ to the power of zero:
First, $\sin {x}^{0} = \sin 1$
so it's integration will be easy with the power rule:
$\int \sin 1 \mathrm{dx}$=$\left(\sin 1\right) \cdot x + C$
.

If you mean $x$ degree then this will be the answer:
${x}_{\mathrm{de} g r e e}$=$\frac{180}{\pi} {x}_{r a \mathrm{di} a n}$
$\int \sin \left({x}^{0}\right) \cdot \mathrm{dx} = \int \sin \left(\frac{180}{\pi} x\right) \cdot \mathrm{dx}$
=$- \cos \frac{\frac{180}{\pi} x}{\frac{180}{\pi}} + C$
=$- \frac{\pi}{180} \cos \left(\frac{180}{\pi} x\right) + C$