What is the integral of sin(3x)sin(7x) ?

The integral of sin(3x)sin(7x) is 1/2 (1/4 sin(4x)-1/10 sin(10x))+C

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integral of sin(3x)sin(7x)

\int sin (3 x) sin (7 x) d x

\frac{1}{2} (\frac{1}{4} sin (4 x) - \frac{1}{10} sin (10 x)) + C

Solution steps

\int sin (3 x) sin (7 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (- cos (10 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 (10 x) + cos (- 4 x) d x

Simplify

- cos (10 x) + cos (- 4 x) : cos (4 x) - cos (10 x)

= \frac{1}{2} \cdot \int cos (4 x) - cos (10 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 (4 x) d x - \int cos (10 x) d x)

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

\int cos (10 x) d x = \frac{1}{10} sin (10 x)

= \frac{1}{2} (\frac{1}{4} sin (4 x) - \frac{1}{10} sin (10 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{4} sin (4 x) - \frac{1}{10} sin (10 x)) + C

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

x \to 1 lim (\frac{x ^{3} + 1}{x + 1})

y^{^{'}} = \frac{x ^{2} + 3 x + 2}{y - 2}

\frac{d}{d x} (arctan^{2} (x^{3}))

\frac{\partial}{\partial y} (e^{y} \cdot sin (x))

What is the integral of sin(3x)sin(7x) ?
The integral of sin(3x)sin(7x) is 1/2 (1/4 sin(4x)-1/10 sin(10x))+C