What is the integral of sin(4x)cos(2x) ?

The integral of sin(4x)cos(2x) is 1/2 (-1/6 cos(6x)-1/2 cos(2x))+C

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integral of sin(4x)cos(2x)

\int sin (4 x) cos (2 x) d x

\frac{1}{2} (- \frac{1}{6} cos (6 x) - \frac{1}{2} cos (2 x)) + C

Solution steps

\int sin (4 x) cos (2 x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int sin (4 x + 2 x) + sin (4 x - 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

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

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

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

= \frac{1}{2} (- \frac{1}{6} cos (6 x) - \frac{1}{2} cos (2 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{6} cos (6 x) - \frac{1}{2} cos (2 x)) + C

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

\frac{9 cos ( x )}{1 + sin ( x )}

\frac{1}{s - 3} - \frac{16}{s ^{2} + 9}

x \to 6 lim (ln (x - 6))

y = (3 x^{2} + 4) (3 + x^{3})^{5}

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