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

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

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

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

\frac{1}{4} cos (4 x) + x + C

Solution steps

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

Apply u-substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

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

\int 1 d u = u

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

Substitute back

u = 2 x

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

Simplify

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

= \frac{1}{4} cos (4 x) + x

Add a constant to the solution

= \frac{1}{4} cos (4 x) + x + C

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