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

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

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integral of cos(4x)sin(x)

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

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

Solution steps

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

Use the following identity:

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

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

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

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

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

Add a constant to the solution

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

y = \frac{x ^{2} + 1}{x ^{3} - 1} - 5 x^{2}

\frac{e ^{x}}{1 + e ^{x}}

x^{3} + 4 y^{2} = 12, (2, - 1)

\frac{d y}{d x} = x - \frac{y}{x}

y^{^{'}} + 5 y = 4

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