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

The integral of (1-sin(2x))^2 is 3/2 x-1/8 sin(4x)+cos(2x)+C

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integral of (1-sin(2x))^2

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

\frac{3}{2} x - \frac{1}{8} sin (4 x) + cos (2 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2} - sin (u) + \frac{1}{2} sin^{2} (u) d u

= \int \frac{1}{2} - sin (u) + \frac{1}{2} sin^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{2} d u - \int sin (u) d u + \int \frac{1}{2} sin^{2} (u) d u

\int \frac{1}{2} d u = \frac{1}{2} u

\int sin (u) d u = - cos (u)

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

= \frac{1}{2} u - (- cos (u)) + \frac{1}{4} (u - \frac{1}{2} sin (2 u))

Substitute back

u = 2 x

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

Simplify

\frac{1}{2} \cdot 2 x - (- cos (2 x)) + \frac{1}{4} (2 x - \frac{1}{2} sin (2 \cdot 2 x)) : \frac{3}{2} x - \frac{1}{8} sin (4 x) + cos (2 x)

= \frac{3}{2} x - \frac{1}{8} sin (4 x) + cos (2 x)

Add a constant to the solution

= \frac{3}{2} x - \frac{1}{8} sin (4 x) + cos (2 x) + C

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What is the integral of (1-sin(2x))^2 ?
The integral of (1-sin(2x))^2 is 3/2 x-1/8 sin(4x)+cos(2x)+C