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)

Add a constant to the solution

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

\frac{d}{d x} (arctan (\frac{( x - 1 )}{x + 1}))

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

\frac{d}{d x} (2 x + 8)

e^{4 x} \cdot 4

π^{x}

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