What is the integral of 30sin(4x-2)cos(x+1) ?

The integral of 30sin(4x-2)cos(x+1) is 15(-1/5 cos(5x-1)-1/3 cos(3x-3))+C

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integral of 30sin(4x-2)cos(x+1)

\int 30 sin (4 x - 2) cos (x + 1) d x

15 (- \frac{1}{5} cos (5 x - 1) - \frac{1}{3} cos (3 x - 3)) + C

Solution steps

\int 30 sin (4 x - 2) cos (x + 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 30 \cdot \int sin (4 x - 2) cos (x + 1) d x

Use the following identity:

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

= 30 \cdot \int \frac{sin ( 4 x - 2 + x + 1 ) + sin ( 4 x - 2 - ( x + 1 ) )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 30 \cdot \frac{1}{2} \cdot \int sin (4 x - 2 + x + 1) + sin (4 x - 2 - (x + 1)) 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

= 30 \cdot \frac{1}{2} (\int sin (4 x - 2 + x + 1) d x + \int sin (4 x - 2 - (x + 1)) d x)

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

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

= 30 \cdot \frac{1}{2} (- \frac{1}{5} cos (5 x - 1) - \frac{1}{3} cos (3 x - 3))

Simplify

30 \cdot \frac{1}{2} (- \frac{1}{5} cos (5 x - 1) - \frac{1}{3} cos (3 x - 3)) : 15 (- \frac{1}{5} cos (5 x - 1) - \frac{1}{3} cos (3 x - 3))

= 15 (- \frac{1}{5} cos (5 x - 1) - \frac{1}{3} cos (3 x - 3))

Add a constant to the solution

= 15 (- \frac{1}{5} cos (5 x - 1) - \frac{1}{3} cos (3 x - 3)) + C

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What is the integral of 30sin(4x-2)cos(x+1) ?
The integral of 30sin(4x-2)cos(x+1) is 15(-1/5 cos(5x-1)-1/3 cos(3x-3))+C