Solve ∫ (from 0 to 2 pi) of (1+sin(x)) wrt x

$$\int _ { 0 } ^ { 2 \pi } ( 1 + \sin ( x ) ) d x$$

Evaluate

$$2\pi \approx 6.283185307$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 1+\sin(x)\mathrm{d}x$$

Integrate the sum term by term.

$$\int 1\mathrm{d}x+\int \sin(x)\mathrm{d}x$$

Find the integral of 1 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$x+\int \sin(x)\mathrm{d}x$$

Use \int \sin(x)\mathrm{d}x=-\cos(x) from the table of common integrals to obtain the result.

$$x-\cos(x)$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$2\pi -\cos(2\pi )+\cos(0)$$

Simplify.

$$2\pi$$