What is the integral of sin(10x)cos(3x) ?

The integral of sin(10x)cos(3x) is 1/2 (-1/13 cos(13x)-1/7 cos(7x))+C

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integral of sin(10x)cos(3x)

\int sin (10 x) cos (3 x) d x

\frac{1}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{7} cos (7 x)) + C

Solution steps

\int sin (10 x) cos (3 x) d x

Use the following identity:

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

= \int \frac{sin ( 10 x + 3 x ) + sin ( 10 x - 3 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 (10 x + 3 x) + sin (10 x - 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 (10 x + 3 x) d x + \int sin (10 x - 3 x) d x)

\int sin (10 x + 3 x) d x = - \frac{1}{13} cos (13 x)

\int sin (10 x - 3 x) d x = - \frac{1}{7} cos (7 x)

= \frac{1}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{7} cos (7 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{7} cos (7 x)) + C

\frac{d}{d x} (9 x + 2)

\int \frac{x + 2 x - 1}{2 x x - 1} d x

\frac{( x ^{3} - 3 )}{( x ^{2} + x - 1 )}

(2 + x^{2}) y^{^{'}} - 2 x y = 0

y = 8 x - x^{2}

What is the integral of sin(10x)cos(3x) ?
The integral of sin(10x)cos(3x) is 1/2 (-1/13 cos(13x)-1/7 cos(7x))+C