What is the integral of (sin(5x)sin(x)) ?

The integral of (sin(5x)sin(x)) is 1/2 (-1/6 sin(6x)+1/4 sin(4x))+C

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integral of (sin(5x)sin(x))

\int (sin (5 x) sin (x)) d x

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

Solution steps

\int sin (5 x) sin (x) d x

Rewrite using trig identities

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

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

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

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

Add a constant to the solution

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

1

\int \frac{x}{2 ( x + 1 ) 2 x + 1} d x

\frac{d}{d x} (\frac{x ^{3}}{3} - \frac{3}{x ^{5}} + 4 π)

\frac{d y}{d x} = \frac{( 4 y lo g _{e} ( x ) )}{x}

\frac{1}{( 1 + x ^{2} ) , ( 1 , \frac{1}{2} )}

What is the integral of (sin(5x)sin(x)) ?
The integral of (sin(5x)sin(x)) is 1/2 (-1/6 sin(6x)+1/4 sin(4x))+C