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

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

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

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

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

Solution steps

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

Expand

(sin (2 x) - cos (2 x))^{2} : sin (2 x) - 2 cos (2 x) sin (2 x) + cos^{2} (2 x)

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

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

\int 2 cos (2 x) sin (2 x) d x = \frac{2}{3} sin^{\frac{3}{2}} (2 x)

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

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

Add a constant to the solution

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

4 x \frac{d y}{d x} + y = x^{3}

\frac{\partial}{\partial x} (e^{6 x e^{y}})

\frac{d y}{d t} = - 2 t y^{2}

\int_{0}^{2} x^{2} e^{x^{3} + 0} d x

\frac{d}{d x} (\frac{3 x}{4})

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