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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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

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

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

\frac{d x}{d y}, y^{2} - x = 0

f (x) = x^{4} + 8 x^{3} - 2 x - 2, at x = - 2

\int \frac{sec ^{2} ( x )}{1 + tan ( x )} d x

\frac{( e ^{x} + e ^{- x} ) d y}{d x} = y^{2}

\frac{d}{d x} (\frac{1}{x ^{2} - 9})

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