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

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

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

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

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

Solution steps

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

Expand

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

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

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

\int 2 sin (x) cos (2 x) d x = - \frac{1}{3} cos (3 x) + cos (x)

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

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

Simplify

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

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

Add a constant to the solution

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

\int (7 + 7 tan^{2} (θ)) d θ

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

f (x) = \frac{x}{( 2 x - 1 ) ^{6}}, at x = 1

\frac{d y}{d x} + 3 y + 5 = 0

cos (2 x)

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