What is the integral of sin(20x)cos(17x) ?

The integral of sin(20x)cos(17x) is 1/2 (-1/37 cos(37x)-1/3 cos(3x))+C

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integral of sin(20x)cos(17x)

\int sin (20 x) cos (17 x) d x

\frac{1}{2} (- \frac{1}{37} cos (37 x) - \frac{1}{3} cos (3 x)) + C

Solution steps

\int sin (20 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 ( 20 x + 17 x ) + sin ( 20 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 (20 x + 17 x) + sin (20 x - 17 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (37 x) + sin (3 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 (37 x) d x + \int sin (3 x) d x)

\int sin (37 x) d x = - \frac{1}{37} cos (37 x)

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

= \frac{1}{2} (- \frac{1}{37} cos (37 x) - \frac{1}{3} cos (3 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{37} cos (37 x) - \frac{1}{3} cos (3 x)) + C

x^{2} y^{^{'}} = y^{2} + 5 x y + 4 x^{2}

\frac{d}{d x} ((4 x^{2} - 8 x + 8) e^{- 6 x})

x = t sin (t)

8 e^{t}

x \to 0 - lim (x^{2} sin (\frac{1}{x}))

What is the integral of sin(20x)cos(17x) ?
The integral of sin(20x)cos(17x) is 1/2 (-1/37 cos(37x)-1/3 cos(3x))+C