What is the integral from 6 to 136 of sin^2(x) ?

The integral from 6 to 136 of sin^2(x) is 1/2 (130-1/2 (sin(272)-sin(12)))

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integral from 6 to 136 of sin^2(x)

\int_{6}^{136} sin^{2} (x) d x

\frac{1}{2} (130 - \frac{1}{2} (sin (272) - sin (12)))

Decimal Notation

64.62376 \dots

Solution steps

\int_{6}^{136} sin^{2} (x) d x

Rewrite using trig identities

= \int_{6}^{136} \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_{6}^{136} 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_{6}^{136} 1 d x - \int_{6}^{136} cos (2 x) d x)

\int_{6}^{136} 1 d x = 130

\int_{6}^{136} cos (2 x) d x = \frac{1}{2} (sin (272) - sin (12))

= \frac{1}{2} (130 - \frac{1}{2} (sin (272) - sin (12)))

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What is the integral from 6 to 136 of sin^2(x) ?
The integral from 6 to 136 of sin^2(x) is 1/2 (130-1/2 (sin(272)-sin(12)))