What is the integral from 0 to a of sin^2(kx) ?

The integral from 0 to a of sin^2(kx) is 1/2 (a-(sin(2ak))/(2k))

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integral from 0 to a of sin^2(kx)

\int_{0}^{a} sin^{2} (k x) d x

\frac{1}{2} (a - \frac{sin ( 2 ak )}{2 k})

Solution steps

\int_{0}^{a} sin^{2} (k x) d x

Rewrite using trig identities

= \int_{0}^{a} \frac{1 - cos ( 2 k 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}^{a} 1 - cos (2 k 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}^{a} 1 d x - \int_{0}^{a} cos (2 k x) d x)

\int_{0}^{a} 1 d x = a

\int_{0}^{a} cos (2 k x) d x = \frac{sin ( 2 ak )}{2 k}

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

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

\int_{2}^{4} (x^{3} - 2) - ln (x + 1) d x

\int_{- 2}^{- 1} \frac{1}{x ^{3}} + \frac{x ^{3}}{4} d x

\int_{3}^{4} (2 x - 5)^{4} d x

\int_{0}^{6} 3 x + 7 d x

What is the integral from 0 to a of sin^2(kx) ?
The integral from 0 to a of sin^2(kx) is 1/2 (a-(sin(2ak))/(2k))