What is the integral from-0.25 to 0.25 of (cos(x))^2 ?

The integral from-0.25 to 0.25 of (cos(x))^2 is 0.48971…

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integral from-0.25 to 0.25 of (cos(x))^2

\int_{- 0.25}^{0.25} (cos (x))^{2} d x

0.48971 \dots

Solution steps

\int_{- 0.25}^{0.25} (cos (x))^{2} d x

Simplify

= \int_{- 0.25}^{0.25} cos^{2} (x) d x

Rewrite using trig identities

= \int_{- 0.25}^{0.25} \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_{- 0.25}^{0.25} 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_{- 0.25}^{0.25} 1 d x + \int_{- 0.25}^{0.25} cos (2 x) d x)

\int_{- 0.25}^{0.25} 1 d x = 0.5

\int_{- 0.25}^{0.25} cos (2 x) d x = 0.47942 \dots

= \frac{1}{2} (0.5 + 0.47942 \dots)

Simplify

\frac{1}{2} (0.5 + 0.47942 \dots) : 0.48971 \dots

= 0.48971 \dots

What is the integral from-0.25 to 0.25 of (cos(x))^2 ?
The integral from-0.25 to 0.25 of (cos(x))^2 is 0.48971…