What is the integral of sin^2(kx) ?

The integral of sin^2(kx) is 1/2 (x-1/(2k)sin(2kx))+C

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integral of sin^2(kx)

\int sin^{2} (k x) d x

\frac{1}{2} (x - \frac{1}{2 k} sin (2 k x)) + C

Solution steps

\int sin^{2} (k x) d x

Rewrite using trig identities

= \int \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 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 1 d x - \int cos (2 k x) d x)

\int 1 d x = x

\int cos (2 k x) d x = \frac{1}{2 k} sin (2 k x)

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

Add a constant to the solution

= \frac{1}{2} (x - \frac{1}{2 k} sin (2 k x)) + C

x \to - 7 - lim (\frac{1}{x ^{3} + 7 x ^{2}})

x \to - \infty lim (9 x^{2})

n = 1 \sum \infty \frac{n}{n ^{2} + n + 1}

\frac{d}{d x} ((3 x - 1)^{4} \cdot (2 x + 1)^{- 3})

x \to 1 lim (\frac{5 - 5 x}{1 - x})

What is the integral of sin^2(kx) ?
The integral of sin^2(kx) is 1/2 (x-1/(2k)sin(2kx))+C