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

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

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integral of 1-sin^2(2x)

\int 1 - sin^{2} (2 x) d x

\frac{1}{2} x + \frac{1}{8} sin (4 x) + C

Solution steps

\int 1 - sin^{2} (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} (2 x) d x

\int 1 d x = x

\int sin^{2} (2 x) d x = \frac{1}{2} (x - \frac{1}{4} sin (4 x))

= x - \frac{1}{2} (x - \frac{1}{4} sin (4 x))

Simplify

x - \frac{1}{2} (x - \frac{1}{4} sin (4 x)) : \frac{1}{2} x + \frac{1}{8} sin (4 x)

= \frac{1}{2} x + \frac{1}{8} sin (4 x)

Add a constant to the solution

= \frac{1}{2} x + \frac{1}{8} sin (4 x) + C

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