What is the integral of-4sin(5x)sin(x) ?

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

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integral of-4sin(5x)sin(x)

\int - 4 sin (5 x) sin (x) d x

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

Solution steps

\int - 4 sin (5 x) sin (x) d x

Take the constant out:

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

= - 4 \cdot \int sin (5 x) sin (x) d x

Rewrite using trig identities

= - 4 \cdot \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

= - 4 \cdot \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

= - 4 \cdot \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)

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

Simplify

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

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

Add a constant to the solution

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

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