What is the integral from a to b of sin^2(x) ?

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

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integral from a to b of sin^2(x)

\int_{a}^{b} sin^{2} (x) d x

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

Solution steps

\int_{a}^{b} sin^{2} (x) d x

Rewrite using trig identities

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

\int_{a}^{b} 1 d x = b - a

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

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

\int_{0}^{4} x^{2} 1 + 2 x d x

\int_{0}^{4} \frac{1}{8} x d x

\int_{3}^{7} (∣ x - 4 ∣ - ∣ x - 5 ∣) d x

\int_{0}^{1} x^{8} (1 + 2 x^{9})^{5} d x

\int_{- \frac{π}{2}}^{π} cos (3 x) d x

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