What is the integral of sin(4x+3)cos(3x-5) ?

The integral of sin(4x+3)cos(3x-5) is 1/2 (-1/7 cos(7x-2)-cos(x+8))+C

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integral of sin(4x+3)cos(3x-5)

\int sin (4 x + 3) cos (3 x - 5) d x

\frac{1}{2} (- \frac{1}{7} cos (7 x - 2) - cos (x + 8)) + C

Solution steps

\int sin (4 x + 3) cos (3 x - 5) d x

Use the following identity:

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

= \int \frac{sin ( 4 x + 3 + 3 x - 5 ) + sin ( 4 x + 3 - ( 3 x - 5 ) )}{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 + 3 x - 5) + sin (4 x + 3 - (3 x - 5)) 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 + 3 + 3 x - 5) d x + \int sin (4 x + 3 - (3 x - 5)) d x)

\int sin (4 x + 3 + 3 x - 5) d x = - \frac{1}{7} cos (7 x - 2)

\int sin (4 x + 3 - (3 x - 5)) d x = - cos (x + 8)

= \frac{1}{2} (- \frac{1}{7} cos (7 x - 2) - cos (x + 8))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{7} cos (7 x - 2) - cos (x + 8)) + C

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What is the integral of sin(4x+3)cos(3x-5) ?
The integral of sin(4x+3)cos(3x-5) is 1/2 (-1/7 cos(7x-2)-cos(x+8))+C