What is the integral of sin(24x)cos(17x) ?

The integral of sin(24x)cos(17x) is 1/2 (-1/41 cos(41x)-1/7 cos(7x))+C

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integral of sin(24x)cos(17x)

\int sin (24 x) cos (17 x) d x

\frac{1}{2} (- \frac{1}{41} cos (41 x) - \frac{1}{7} cos (7 x)) + C

Solution steps

\int sin (24 x) cos (17 x) d x

Use the following identity:

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

= \int \frac{sin ( 24 x + 17 x ) + sin ( 24 x - 17 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 (24 x + 17 x) + sin (24 x - 17 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (41 x) + sin (7 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 (41 x) d x + \int sin (7 x) d x)

\int sin (41 x) d x = - \frac{1}{41} cos (41 x)

\int sin (7 x) d x = - \frac{1}{7} cos (7 x)

= \frac{1}{2} (- \frac{1}{41} cos (41 x) - \frac{1}{7} cos (7 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{41} cos (41 x) - \frac{1}{7} cos (7 x)) + C

\int x - \frac{x ^{2}}{2} + c d x

n = 1 \sum \infty \frac{1}{n ^{2} + 6 n + 8}

x \to 0 lim (\frac{( cos ( x ) )}{x})

\frac{d}{d x} (\frac{1}{1 + 4 x ^{2}})

\frac{d}{d t} (π t)

What is the integral of sin(24x)cos(17x) ?
The integral of sin(24x)cos(17x) is 1/2 (-1/41 cos(41x)-1/7 cos(7x))+C