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

The integral from 0 to T of sin^2(x) is 1/2 (T-1/2 sin(2T))

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integral from 0 to T of sin^2(x)

\int_{0}^{T} sin^{2} (x) d x

\frac{1}{2} (T - \frac{1}{2} sin (2 T))

Solution steps

\int_{0}^{T} sin^{2} (x) d x

Rewrite using trig identities

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

\int_{0}^{T} 1 d x = T

\int_{0}^{T} cos (2 x) d x = \frac{1}{2} sin (2 T)

= \frac{1}{2} (T - \frac{1}{2} sin (2 T))

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