What is the integral of sin(2x)sin(4x)sin(6x) ?

The integral of sin(2x)sin(4x)sin(6x) is 1/4 (-1/2 (-1/6 cos(12x)+1/4 cos(8x))+1/2 sin^2(2x))+C

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integral of sin(2x)sin(4x)sin(6x)

\int sin (2 x) sin (4 x) sin (6 x) d x

\frac{1}{4} (- \frac{1}{2} (- \frac{1}{6} cos (12 x) + \frac{1}{4} cos (8 x)) + \frac{1}{2} sin^{2} (2 x)) + C

Solution steps

\int sin (2 x) sin (4 x) sin (6 x) d x

Apply u-substitution

: \int \frac{1}{2} sin (u) sin (2 u) sin (3 u) d u

= \int \frac{1}{2} sin (u) sin (2 u) sin (3 u) d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sin (u) sin (2 u) sin (3 u) d u

Rewrite using trig identities

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

Take the constant out:

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

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

Expand

sin (u) (- cos (5 u) + cos (- u)) : - sin (u) cos (5 u) + sin (u) cos (- u)

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

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

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

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

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

Substitute back

u = 2 x

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

Simplify

\frac{1}{2} \cdot \frac{1}{2} (- \frac{1}{2} (- \frac{1}{6} cos (6 \cdot 2 x) + \frac{1}{4} cos (4 \cdot 2 x)) + \frac{sin ^{2} ( 2 x )}{2}) : \frac{1}{4} (- \frac{1}{2} (- \frac{1}{6} cos (12 x) + \frac{1}{4} cos (8 x)) + \frac{1}{2} sin^{2} (2 x))

= \frac{1}{4} (- \frac{1}{2} (- \frac{1}{6} cos (12 x) + \frac{1}{4} cos (8 x)) + \frac{1}{2} sin^{2} (2 x))

Add a constant to the solution

= \frac{1}{4} (- \frac{1}{2} (- \frac{1}{6} cos (12 x) + \frac{1}{4} cos (8 x)) + \frac{1}{2} sin^{2} (2 x)) + C

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What is the integral of sin(2x)sin(4x)sin(6x) ?
The integral of sin(2x)sin(4x)sin(6x) is 1/4 (-1/2 (-1/6 cos(12x)+1/4 cos(8x))+1/2 sin^2(2x))+C