What is the integral of sin(4x)cos(17x) ?

The integral of sin(4x)cos(17x) is 1/2 (-1/21 cos(21x)+1/13 cos(13x))+C

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integral of sin(4x)cos(17x)

\int sin (4 x) cos (17 x) d x

\frac{1}{2} (- \frac{1}{21} cos (21 x) + \frac{1}{13} cos (13 x)) + C

Solution steps

\int sin (4 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 ( 4 x + 17 x ) + sin ( 4 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 (4 x + 17 x) + sin (4 x - 17 x) d x

Simplify

sin (4 x + 17 x) + sin (4 x - 17 x) : sin (21 x) - sin (13 x)

= \frac{1}{2} \cdot \int sin (21 x) - sin (13 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 (21 x) d x - \int sin (13 x) d x)

\int sin (21 x) d x = - \frac{1}{21} cos (21 x)

\int sin (13 x) d x = - \frac{1}{13} cos (13 x)

= \frac{1}{2} (- \frac{1}{21} cos (21 x) - (- \frac{1}{13} cos (13 x)))

Simplify

= \frac{1}{2} (- \frac{1}{21} cos (21 x) + \frac{1}{13} cos (13 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{21} cos (21 x) + \frac{1}{13} cos (13 x)) + C

n = 0 \sum \infty (\frac{2}{11})^{n}

- 6 x^{2} + 5 x

y = (ln (x))^{3}

\int \frac{x + 3}{x - 1} d x

f (x) = e^{2 - x}

What is the integral of sin(4x)cos(17x) ?
The integral of sin(4x)cos(17x) is 1/2 (-1/21 cos(21x)+1/13 cos(13x))+C