What is the integral of (sin(2x)-cos(3x))^2 ?

The integral of (sin(2x)-cos(3x))^2 is x-1/8 sin(4x)+1/5 cos(5x)+1/12 sin(6x)-cos(x)+C

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integral of (sin(2x)-cos(3x))^2

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

x - \frac{1}{8} sin (4 x) + \frac{1}{5} cos (5 x) + \frac{1}{12} sin (6 x) - cos (x) + C

Solution steps

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

Factor

sin (2 x) - cos (3 x) : - (- sin (2 x) + cos (3 x))

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

Simplify

(- (- sin (2 x) + cos (3 x)))^{2} : (- sin (2 x) + cos (3 x))^{2}

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

Expand

(- sin (2 x) + cos (3 x))^{2} : sin^{2} (2 x) - 2 sin (2 x) cos (3 x) + cos^{2} (3 x)

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

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

\int 2 sin (2 x) cos (3 x) d x = - \frac{1}{5} cos (5 x) + cos (x)

\int cos^{2} (3 x) d x = \frac{1}{2} (x + \frac{1}{6} sin (6 x))

= \frac{1}{2} (x - \frac{1}{4} sin (4 x)) - (- \frac{1}{5} cos (5 x) + cos (x)) + \frac{1}{2} (x + \frac{1}{6} sin (6 x))

Simplify

\frac{1}{2} (x - \frac{1}{4} sin (4 x)) - (- \frac{1}{5} cos (5 x) + cos (x)) + \frac{1}{2} (x + \frac{1}{6} sin (6 x)) : x - \frac{1}{8} sin (4 x) + \frac{1}{5} cos (5 x) + \frac{1}{12} sin (6 x) - cos (x)

= x - \frac{1}{8} sin (4 x) + \frac{1}{5} cos (5 x) + \frac{1}{12} sin (6 x) - cos (x)

Add a constant to the solution

= x - \frac{1}{8} sin (4 x) + \frac{1}{5} cos (5 x) + \frac{1}{12} sin (6 x) - cos (x) + C

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What is the integral of (sin(2x)-cos(3x))^2 ?
The integral of (sin(2x)-cos(3x))^2 is x-1/8 sin(4x)+1/5 cos(5x)+1/12 sin(6x)-cos(x)+C