What is the integral of 3sin(2x)cos(3x) ?

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

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integral of 3sin(2x)cos(3x)

\int 3 sin (2 x) cos (3 x) d x

\frac{3}{2} (- \frac{1}{5} cos (5 x) + cos (x)) + C

Solution steps

\int 3 sin (2 x) cos (3 x) d x

Take the constant out:

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

= 3 \cdot \int sin (2 x) cos (3 x) d x

Use the following identity:

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

= 3 \cdot \int \frac{sin ( 2 x + 3 x ) + sin ( 2 x - 3 x )}{2} d x

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int sin (2 x + 3 x) + sin (2 x - 3 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

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

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

\int sin (2 x - 3 x) d x = cos (x)

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

Simplify

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

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

Add a constant to the solution

= \frac{3}{2} (- \frac{1}{5} cos (5 x) + cos (x)) + C

\int_{0}^{4} ((4 - x) x) d x

\frac{d y}{d x} = e^{4 x} - 3 y

x \to \infty lim (x^{2} y^{2})

f (x) = \frac{3 x - 9}{2 x + 8}

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

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