What is the integral of sin(7x)sin(4x) ?

The integral of sin(7x)sin(4x) is 1/2 (-1/11 sin(11x)+1/3 sin(3x))+C

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integral of sin(7x)sin(4x)

\int sin (7 x) sin (4 x) d x

\frac{1}{2} (- \frac{1}{11} sin (11 x) + \frac{1}{3} sin (3 x)) + C

Solution steps

\int sin (7 x) sin (4 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (- cos (11 x) + cos (3 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 (11 x) + cos (3 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 (11 x) d x + \int cos (3 x) d x)

\int cos (11 x) d x = \frac{1}{11} sin (11 x)

\int cos (3 x) d x = \frac{1}{3} sin (3 x)

= \frac{1}{2} (- \frac{1}{11} sin (11 x) + \frac{1}{3} sin (3 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{11} sin (11 x) + \frac{1}{3} sin (3 x)) + C

f (x) = 6 x^{2} + 8 x - 15, at x = 2

\frac{\partial}{\partial y} (e^{- x^{2} - y^{2}})

x \to 3 lim (\frac{tan ( x - 3 )}{3 e ^{x - 3} - x})

\frac{1}{x ^{2} + 4}

\frac{d}{d x} (x^{7} ln (8 x))

What is the integral of sin(7x)sin(4x) ?
The integral of sin(7x)sin(4x) is 1/2 (-1/11 sin(11x)+1/3 sin(3x))+C