What is the integral of (sin(2u)-cos(3u))^{(2)} ?

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

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integral of (sin(2u)-cos(3u))^{(2)}

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

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

Solution steps

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

Factor

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

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

Simplify

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

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

Expand

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

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (- 3 x - 9)

\int \frac{y}{( 1 + y ^{2} ) ^{4}} d y

[x + (x + sin^{2} (x))^{3}]^{4}

x \to 2 + lim (\frac{x ^{2} - 4}{x - 2})

\int tan^{5} (4 x) sec^{4} (4 x) d x

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