What is the integral of sin(16x)cos(11x) ?

The integral of sin(16x)cos(11x) is 1/2 (-1/27 cos(27x)-1/5 cos(5x))+C

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integral of sin(16x)cos(11x)

\int sin (16 x) cos (11 x) d x

\frac{1}{2} (- \frac{1}{27} cos (27 x) - \frac{1}{5} cos (5 x)) + C

Solution steps

\int sin (16 x) cos (11 x) d x

Use the following identity:

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

= \int \frac{sin ( 16 x + 11 x ) + sin ( 16 x - 11 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 (16 x + 11 x) + sin (16 x - 11 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (27 x) + sin (5 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 (27 x) d x + \int sin (5 x) d x)

\int sin (27 x) d x = - \frac{1}{27} cos (27 x)

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

= \frac{1}{2} (- \frac{1}{27} cos (27 x) - \frac{1}{5} cos (5 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{27} cos (27 x) - \frac{1}{5} cos (5 x)) + C

\frac{d N}{d t} = 5 - N

\frac{2 x + x ^{2}}{( 1 + x ) ^{2}}

\frac{d y}{d x} = \frac{x - y}{x + 2 y}

(9 + x) \frac{d y}{d x} = 6 + x

\frac{d}{d x} (\frac{x y}{2})

What is the integral of sin(16x)cos(11x) ?
The integral of sin(16x)cos(11x) is 1/2 (-1/27 cos(27x)-1/5 cos(5x))+C