What is the integral of sin(7x)sin(12x) ?

The integral of sin(7x)sin(12x) is 1/2 (1/5 sin(5x)-1/19 sin(19x))+C

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integral of sin(7x)sin(12x)

\int sin (7 x) sin (12 x) d x

\frac{1}{2} (\frac{1}{5} sin (5 x) - \frac{1}{19} sin (19 x)) + C

Solution steps

\int sin (7 x) sin (12 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (- cos (19 x) + cos (- 5 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 (19 x) + cos (- 5 x) d x

Simplify

- cos (19 x) + cos (- 5 x) : cos (5 x) - cos (19 x)

= \frac{1}{2} \cdot \int cos (5 x) - cos (19 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 (5 x) d x - \int cos (19 x) d x)

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

\int cos (19 x) d x = \frac{1}{19} sin (19 x)

= \frac{1}{2} (\frac{1}{5} sin (5 x) - \frac{1}{19} sin (19 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{5} sin (5 x) - \frac{1}{19} sin (19 x)) + C

\frac{d}{d x} (\frac{x ^{3}}{24} + \frac{2}{x})

\frac{d}{d x} (cos (x y) - 3 y^{4} + 3)

\frac{\partial}{\partial x} (\frac{A}{x + 5 x})

\frac{d}{d x} ((1 + x) (x^{3}))

\frac{\partial}{\partial y} (x^{\frac{1}{3}} y^{\frac{2}{3}})

What is the integral of sin(7x)sin(12x) ?
The integral of sin(7x)sin(12x) is 1/2 (1/5 sin(5x)-1/19 sin(19x))+C