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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\int_{0}^{1} \frac{y}{2} d y

\int_{0}^{1} (6 t)^{2} + (6 t^{2})^{2} d t

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))