What is the integral from 2 to 6 of cos^2(x) ?

The integral from 2 to 6 of cos^2(x) is 1/2 (4+1/2 (sin(12)-sin(4)))

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integral from 2 to 6 of cos^2(x)

\int_{2}^{6} cos^{2} (x) d x

\frac{1}{2} (4 + \frac{1}{2} (sin (12) - sin (4)))

Decimal

2.05505 \dots

Solution steps

\int_{2}^{6} cos^{2} (x) d x

Rewrite using trig identities

= \int_{2}^{6} \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_{2}^{6} 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_{2}^{6} 1 d x + \int_{2}^{6} cos (2 x) d x)

\int_{2}^{6} 1 d x = 4

\int_{2}^{6} cos (2 x) d x = \frac{1}{2} (sin (12) - sin (4))

= \frac{1}{2} (4 + \frac{1}{2} (sin (12) - sin (4)))

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What is the integral from 2 to 6 of cos^2(x) ?
The integral from 2 to 6 of cos^2(x) is 1/2 (4+1/2 (sin(12)-sin(4)))