What is the integral from 0 to 1 of (sin(2pit))^2 ?

The integral from 0 to 1 of (sin(2pit))^2 is 1/2

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integral from 0 to 1 of (sin(2pit))^2

\int_{0}^{1} (sin (2 π t))^{2} d t

\frac{1}{2}

Decimal Notation

0.5

Solution steps

\int_{0}^{1} (sin (2 π t))^{2} d t

Simplify

= \int_{0}^{1} sin^{2} (2 π t) d t

Rewrite using trig identities

= \int_{0}^{1} \frac{1}{2} (1 - cos (4 π t)) 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}^{1} 1 - cos (4 π 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}^{1} 1 d t - \int_{0}^{1} cos (4 π t) d t)

\int_{0}^{1} 1 d t = 1

\int_{0}^{1} cos (4 π t) d t = 0

= \frac{1}{2} (1 - 0)

Simplify

= \frac{1}{2}

\frac{460 t}{( t ^{2} + 5 ) ^{3}}

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\frac{d}{d x} (ln (sin (x) + cos (x)))

\int - x^{2} + 6 x d x

\frac{d}{d x} (arctan (4 x^{2} - 1))

What is the integral from 0 to 1 of (sin(2pit))^2 ?
The integral from 0 to 1 of (sin(2pit))^2 is 1/2