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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

\int cos (6 x) d x = \frac{1}{6} sin (6 x)

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

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

Add a constant to the solution

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

x \to - 1 lim (\frac{x}{x ^{2} - 1})

\frac{x ^{3} + 3 x}{x ^{2} - 4 x + 3}

x \to 4 - lim (\frac{∣ x - 4 ∣}{x - 4})

\int (x^{3} + x) x^{4} + 2 x^{2} d x

\frac{\partial}{\partial x} (sin (x z^{2}) - x y^{2})

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