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

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

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

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

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

Solution steps

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

Use the following identity:

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

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

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

\int sin (3 x - 4 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 \frac{( x ^{2} + 1 )}{( x - 6 ) ( x - 5 ) ^{2}} d x

\int 7 x + 4 d x

\frac{\partial}{\partial x} (3 x y^{6} + sin (π x) - z)

\int x (6 x^{2} - x + 7) d x

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