Question

How do you find the integral of `(sin 2x + 1)^2`?

Answer 1

`int (sin2x+1)^2dx = 3/2 x -cos 2x -1/8sin 4x +C`

Explanation:

Expand the power of the binomial:

`int (sin2x+1)^2dx = int (sin^2 2x + 2 sin 2x +1)dx`

Use the linearity of the integral:

`int (sin2x+1)^2dx = int sin^2 2x dx + 2 int sin 2x dx +int dx`

Solve the integrals separately:

`int dx = x+C_1`

`2 int sin 2x dx = int sin 2x d(2x) = -cos 2x+C_2`

`int sin^2 2x dx = int (1-cos 4x)/2dx = 1/2x -1/8 sin(4x) +C_3`

And collecting the terms:

`int (sin2x+1)^2dx = 3/2 x -cos 2x -1/8sin 4x +C`

and applying trigonometric formulas to reduce to functions of the same angle:

`int (sin2x+1)^2dx = 3/2 x -cos 2x -1/4sin 2x cos 2x+C`

`int (sin2x+1)^2dx = 3/2 x -cos 2x (1/4sin 2x +1)+C`