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

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

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

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

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

Solution steps

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

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int \frac{sin ( 4 x + 7 x ) + sin ( 4 x - 7 x )}{2} 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 sin (4 x + 7 x) + sin (4 x - 7 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 sin (4 x + 7 x) d x + \int sin (4 x - 7 x) d x)

\int sin (4 x + 7 x) d x = - \frac{1}{11} cos (11 x)

\int sin (4 x - 7 x) d x = \frac{1}{3} cos (3 x)

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

Add a constant to the solution

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

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

\int_{0}^{0.5} 150 x d x

\int_{0}^{1} \frac{29}{4 y - 1} d y

x \to 0 lim (3 x^{2} - 1)

y = 2 x^{4} + 3 x - 3

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