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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Rewrite using trig identities

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

Expand

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

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

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

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

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

Add a constant to the solution

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

\int \frac{x}{x - 17} d x

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

(1 + x^{5}) y^{^{'}} + 5 x^{4} y = 13 x^{4}, y (0) = 2

\int 9 sin^{3} (x co) s^{7} x d x

\frac{\partial}{\partial x} (3 x ln (4 y + 1))

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