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

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

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integral of cos(x)sin(3x)

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

\frac{1}{2} (- \frac{1}{4} cos (4 x) - \frac{1}{2} cos (2 x)) + C

Solution steps

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

Use the following identity:

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

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

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

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

= \frac{1}{2} (- \frac{1}{4} cos (4 x) - \frac{1}{2} cos (2 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{4} cos (4 x) - \frac{1}{2} cos (2 x)) + C

\frac{\partial}{\partial y} (x \cdot e^{y})

\frac{d}{d x} (e^{x} - 1)

\int_{0}^{x} 2 t d t

\frac{d}{d x} (- \frac{3}{16 x ^{\frac{7}{4}}})

ln (9 - 2 x)

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