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

The integral from 0 to 2pi of (1+sin(x))^2 is 3pi

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

\int_{0}^{2 π} (1 + sin (x))^{2} d x

3 π

Decimal Notation

9.42477 \dots

Solution steps

\int_{0}^{2 π} (1 + sin (x))^{2} d x

Expand

(1 + sin (x))^{2} : 1 + 2 sin (x) + sin^{2} (x)

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

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} 2 sin (x) d x = 0

\int_{0}^{2 π} sin^{2} (x) d x = π

= 2 π + 0 + π

Simplify

= 3 π

\int_{0}^{1 + i} z^{2} d z

\int_{- 1}^{1} e^{x} \cdot x d x

\int_{0}^{1} (3 x - 3 x^{2}) d x

\int_{0}^{π} (π^{2} - x^{2}) d x

\int_{- π}^{π} cos (2 x) cos (2 x) d x

What is the integral from 0 to 2pi of (1+sin(x))^2 ?
The integral from 0 to 2pi of (1+sin(x))^2 is 3pi