What is the integral of 1-cos(4x) ?

The integral of 1-cos(4x) is x-1/4 sin(4x)+C

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integral of 1-cos(4x)

\int 1 - cos (4 x) d x

x - \frac{1}{4} sin (4 x) + C

Solution steps

\int 1 - cos (4 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

= \int 1 d x - \int cos (4 x) d x

\int 1 d x = x

\int cos (4 x) d x = \frac{1}{4} sin (4 x)

= x - \frac{1}{4} sin (4 x)

Add a constant to the solution

= x - \frac{1}{4} sin (4 x) + C

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

\frac{d}{d x} (2 e^{- 2 x} x - 2 e^{- 2 x} x^{2})

(4 - x^{2})^{^{'}}

\frac{d}{d x} (sin (x) e^{- c o s (x)})

x \to 100000 lim (x^{3} + 2 x + 1)

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