What is the integral of cos(5x)sin(6x) ?

The integral of cos(5x)sin(6x) is 1/2 (-1/11 cos(11x)-cos(x))+C

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integral of cos(5x)sin(6x)

\int cos (5 x) sin (6 x) d x

\frac{1}{2} (- \frac{1}{11} cos (11 x) - cos (x)) + C

Solution steps

\int cos (5 x) sin (6 x) d x

Use the following identity:

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

= \int \frac{sin ( 6 x + 5 x ) + sin ( 6 x - 5 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 (6 x + 5 x) + sin (6 x - 5 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (11 x) + sin (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 (11 x) d x + \int sin (x) d x)

\int sin (11 x) d x = - \frac{1}{11} cos (11 x)

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

= \frac{1}{2} (- \frac{1}{11} cos (11 x) - cos (x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{11} cos (11 x) - cos (x)) + C

\int x x + 3 d x

\frac{d y}{d x} = \frac{( x + 3 y )}{y - 3 x}

x \to 3 lim (\frac{x ^{3} - 6 x ^{2} + 9}{x ^{2} - 9})

y = 7 x - 4 x^{2}, (- 3, - 57)

x \to 999 lim (x^{2} - 1)

What is the integral of cos(5x)sin(6x) ?
The integral of cos(5x)sin(6x) is 1/2 (-1/11 cos(11x)-cos(x))+C