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

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

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integral of sin(3x)sin(2x)sin(x)

\int sin (3 x) sin (2 x) sin (x) d x

\frac{1}{2} (\frac{1}{6} cos^{2} (3 x) + \frac{1}{2} (- \frac{1}{4} cos (4 x) - \frac{1}{2} cos (2 x))) + C

Solution steps

\int sin (3 x) sin (2 x) sin (x) d x

Rewrite using trig identities

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

Expand

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

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

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

\int sin (3 x) cos (x) d x = \frac{1}{2} (- \frac{1}{4} cos (4 x) - \frac{1}{2} cos (2 x))

= \frac{1}{2} (- (- \frac{1}{6} cos^{2} (3 x)) + \frac{1}{2} (- \frac{1}{4} cos (4 x) - \frac{1}{2} cos (2 x)))

Simplify

= \frac{1}{2} (\frac{1}{6} cos^{2} (3 x) + \frac{1}{2} (- \frac{1}{4} cos (4 x) - \frac{1}{2} cos (2 x)))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{6} cos^{2} (3 x) + \frac{1}{2} (- \frac{1}{4} cos (4 x) - \frac{1}{2} cos (2 x))) + C

\frac{1}{s + \frac{1}{a}}

x, \frac{1}{x ^{2}}, 2

\frac{d}{d a} (sin^{4} (a) - cos^{4} (a))

\frac{1}{4} y^{^{''}} + 4 y = 2 tan (4 t) - \frac{1}{3} e^{3 t}

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