What is the integral of sin^2(4x) ?

The integral of sin^2(4x) is 1/2 (x-1/8 sin(8x))+C

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integral of sin^2(4x)

\int sin^{2} (4 x) d x

\frac{1}{2} (x - \frac{1}{8} sin (8 x)) + C

Solution steps

\int sin^{2} (4 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (8 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 (8 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 (8 x) d x)

\int 1 d x = x

\int cos (8 x) d x = \frac{1}{8} sin (8 x)

= \frac{1}{2} (x - \frac{1}{8} sin (8 x))

Add a constant to the solution

= \frac{1}{2} (x - \frac{1}{8} sin (8 x)) + C

\frac{d}{d y} (10 y - y^{2})

\frac{\partial}{\partial y} (x^{y - 1})

x (\frac{d y}{d x}) + 2 y = 3

\frac{d y}{d x} y = 7^{x}

\frac{\partial}{\partial y} (\frac{2}{y})

What is the integral of sin^2(4x) ?
The integral of sin^2(4x) is 1/2 (x-1/8 sin(8x))+C