What is the integral from 0 to 3 of sin^2(x) ?

The integral from 0 to 3 of sin^2(x) is 1/2 (3-1/2 sin(6))

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integral from 0 to 3 of sin^2(x)

\int_{0}^{3} sin^{2} (x) d x

\frac{1}{2} (3 - \frac{1}{2} sin (6))

Solution steps

\int_{0}^{3} sin^{2} (x) d x

Rewrite using trig identities

= \int_{0}^{3} \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}^{3} 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}^{3} 1 d x - \int_{0}^{3} cos (2 x) d x)

\int_{0}^{3} 1 d x = 3

\int_{0}^{3} cos (2 x) d x = \frac{1}{2} sin (6)

= \frac{1}{2} (3 - \frac{1}{2} sin (6))

\int_{0}^{x} \frac{1}{x - 1} d x

\int_{- 2}^{3} x - (x^{2} - 6) d x

\int_{0}^{\infty} x e^{- k x} d x

\int_{1}^{3} x^{2} + 8 d x

\int_{x}^{1} x^{2} u^{2} + 1 d u

What is the integral from 0 to 3 of sin^2(x) ?
The integral from 0 to 3 of sin^2(x) is 1/2 (3-1/2 sin(6))