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

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

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

\int_{0}^{2 π} (- sin^{2} (t)) d t

- π

Decimal

- 3.14159 \dots

Solution steps

\int_{0}^{2 π} - sin^{2} (t) d t

Take the constant out:

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

= - \int_{0}^{2 π} sin^{2} (t) d t

Rewrite using trig identities

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

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 t) d t

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 t - \int_{0}^{2 π} cos (2 t) d t)

\int_{0}^{2 π} 1 d t = 2 π

\int_{0}^{2 π} cos (2 t) d t = 0

= - \frac{1}{2} (2 π - 0)

Simplify

- \frac{1}{2} (2 π - 0) : - π

= - π

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\int_{0}^{1} x^{2} 6 - 2 x^{3} d x

\int_{0}^{y} 2 x yz d z

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