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{- 2 x}{( x ^{2} - 1 ) ^{2}}

(cos (3 x)) y^{^{'}} - 3 (tan (3 x)) y = 2 (sec (3 x))

f (t) = 4 t

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

x \to 0 lim (\frac{x}{ln ( 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