Solve ∫ (from 0 to 2 pi) of (sinθ+cosθ)dθ

$$\int _ { 0 } ^ { 2 \pi } ( \sin \theta + \cos \theta ) d \theta$$

Evaluate

$$0$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \sin(\theta )+\cos(\theta )\mathrm{d}\theta$$

Integrate the sum term by term.

$$\int \sin(\theta )\mathrm{d}\theta +\int \cos(\theta )\mathrm{d}\theta$$

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

$$-\cos(\theta )+\int \cos(\theta )\mathrm{d}\theta$$

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

$$-\cos(\theta )+\sin(\theta )$$

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.

$$-\cos(2\pi )+\sin(2\pi )-\left(-\cos(0)+\sin(0)\right)$$

Simplify.

$$0$$