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

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

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integral from-1 to 1 of (sin(x))^2

\int_{- 1}^{1} (sin (x))^{2} d x

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

Decimal

0.54535 \dots

Solution steps

\int_{- 1}^{1} (sin (x))^{2} d x

Simplify

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

Rewrite using trig identities

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

\int_{- 1}^{1} 1 d x = 2

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

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

Simplify

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

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

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