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

The integral from 0 to 2pi of-sin^2(x) is -pi

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

\int_{0}^{2 π} - sin^{2} (x) d x

- π

Decimal Notation

- 3.14159 \dots

Solution steps

\int_{0}^{2 π} - sin^{2} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int_{0}^{2 π} sin^{2} (x) d x

Rewrite using trig identities

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int_{0}^{2 π} 1 - cos (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

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

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} cos (2 x) d x = 0

= - \frac{1}{2} (2 π - 0)

Simplify

- \frac{1}{2} (2 π - 0) : - π

= - π

s \to 0 lim (\frac{4}{1 + 2 ^{\frac{1}{s}}})

\frac{d}{d x} (\frac{x + 1}{x ^{2} + 1})

\frac{d}{d x} (2 x e^{- 2 x} - 2 x^{2} e^{- 2 x})

(\frac{cos ( x )}{sin ( x )})^{^{'}}

\frac{d}{d x} (∣ x - 3 ∣)

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