What is the integral of sin(3x)cos(7x) ?

The integral of sin(3x)cos(7x) is 1/2 (-1/10 cos(10x)+1/4 cos(4x))+C

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integral of sin(3x)cos(7x)

\int sin (3 x) cos (7 x) d x

\frac{1}{2} (- \frac{1}{10} cos (10 x) + \frac{1}{4} cos (4 x)) + C

Solution steps

\int sin (3 x) cos (7 x) d x

Use the following identity:

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

= \int \frac{sin ( 3 x + 7 x ) + sin ( 3 x - 7 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 (3 x + 7 x) + sin (3 x - 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 (3 x + 7 x) d x + \int sin (3 x - 7 x) d x)

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

\int sin (3 x - 7 x) d x = \frac{1}{4} cos (4 x)

= \frac{1}{2} (- \frac{1}{10} cos (10 x) + \frac{1}{4} cos (4 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{10} cos (10 x) + \frac{1}{4} cos (4 x)) + C

x^{2} y^{^{''}} + 3 x y^{^{'}} - 24 y = 0, y (1) = 3, y^{^{'}} (1) = - 3

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

\int \frac{1}{1 + sin ( θ )} d θ

\int 2 x^{2} - x^{3} d x

\frac{\partial}{\partial x} (\frac{x}{2 x ^{2} + 2 y ^{2} + 8})

What is the integral of sin(3x)cos(7x) ?
The integral of sin(3x)cos(7x) is 1/2 (-1/10 cos(10x)+1/4 cos(4x))+C