What is the integral from 0 to 2pi of (1+cos(x)) ?

The integral from 0 to 2pi of (1+cos(x)) is 2pi

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integral from 0 to 2pi of (1+cos(x))

\int_{0}^{2 π} (1 + cos (x)) d x

2 π

Decimal

6.28318 \dots

Solution steps

\int_{0}^{2 π} 1 + cos (x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{2 π} 1 d x + \int_{0}^{2 π} cos (x) d x

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} cos (x) d x = 0

= 2 π + 0

Simplify

= 2 π

\int_{0.2}^{0.8} 2 x d x

\int_{1}^{2} \frac{1}{x ( ln ( x ) )} d x

\int_{1}^{π} ln (t) d t

\int_{0}^{18} x^{2} d x

\int_{1}^{1} x ln (x) d x

What is the integral from 0 to 2pi of (1+cos(x)) ?
The integral from 0 to 2pi of (1+cos(x)) is 2pi