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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

\int_{0}^{x^{2}} 1 d t = x^{2}

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

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

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