What is the integral of 1-sin(2x) ?

The integral of 1-sin(2x) is x+1/2 cos(2x)+C

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integral of 1-sin(2x)

\int 1 - sin (2 x) d x

x + \frac{1}{2} cos (2 x) + C

Solution steps

\int 1 - sin (2 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 sin (2 x) d x

\int 1 d x = x

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

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

Simplify

= x + \frac{1}{2} cos (2 x)

Add a constant to the solution

= x + \frac{1}{2} cos (2 x) + C

x, \frac{1}{18 x ^{2}}, \frac{x}{16}

\frac{d y}{d x} = 60 y x^{14}

\int \frac{40}{1 - cos ( 5 x )} d x

\int \frac{1}{7 x lo g _{7} ( x )} d x

\frac{5}{2 ( 2 x + 3 ) ^{2}}

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