What is the integral of sin^2(9x) ?

The integral of sin^2(9x) is 1/2 (x-1/18 sin(18x))+C

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integral of sin^2(9x)

\int sin^{2} (9 x) d x

\frac{1}{2} (x - \frac{1}{18} sin (18 x)) + C

Solution steps

\int sin^{2} (9 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (18 x)) 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 (18 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 (18 x) d x)

\int 1 d x = x

\int cos (18 x) d x = \frac{1}{18} sin (18 x)

= \frac{1}{2} (x - \frac{1}{18} sin (18 x))

Add a constant to the solution

= \frac{1}{2} (x - \frac{1}{18} sin (18 x)) + C

\int_{0}^{3} - x + 2 d x

\int 6 x^{5} 3 x^{4} + 2 d x

y = 9 x, y = 70 - x^{2}

\int (\frac{5}{x}) d x

\frac{\partial}{\partial y} (3 x^{2} y - y^{3} + x - y)

What is the integral of sin^2(9x) ?
The integral of sin^2(9x) is 1/2 (x-1/18 sin(18x))+C