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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} ((4 x^{3} - 6) (4 x^{2} + 4))

\int \frac{1}{( 1 - x ^{2} ) ^{3}} d x

\frac{d}{d x} (\frac{ln ( x )}{9 x})

\int (\frac{1}{e ^{3 t}}) d t

x \to 0 lim (\frac{e - 1}{x})

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