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

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

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

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

\frac{\partial}{\partial x} (25 x^{(\frac{3}{4})} y^{\frac{1}{4}})

y^{^{'}} + \frac{2 ( y - 1 )}{x} = 0

\int_{0}^{1} e^{2 x} d x

a re a 15 - x^{2}, x^{2} - 3

n = 1 \sum \infty 2 e^{- 9 n}

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