What is the integral of sin(4x)cos(3x) ?

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

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integral of sin(4x)cos(3x)

\int sin (4 x) cos (3 x) d x

\frac{1}{2} (- \frac{1}{7} cos (7 x) - cos (x)) + C

Solution steps

\int sin (4 x) cos (3 x) d x

Use the following identity:

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

= \int \frac{sin ( 4 x + 3 x ) + sin ( 4 x - 3 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 + 3 x) + sin (4 x - 3 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (7 x) + sin (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 (7 x) d x + \int sin (x) d x)

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

\int sin (x) d x = - cos (x)

= \frac{1}{2} (- \frac{1}{7} cos (7 x) - cos (x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{7} cos (7 x) - cos (x)) + C

\int (x^{2} + 6) e^{- x} d x

\int \frac{7}{x} + 7 x d x

\frac{\partial ^{2}}{\partial x ^{2}} (ln (x + 2 y))

n = 1 \sum \infty \frac{1}{3 n ^{2} + 2}

\frac{d}{d x} (\frac{2 cos ( x ) - 3}{sin ( x )})

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