What is the integral of sin(7x)cos(2x) ?

The integral of sin(7x)cos(2x) is 1/2 (-1/9 cos(9x)-1/5 cos(5x))+C

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integral of sin(7x)cos(2x)

\int sin (7 x) cos (2 x) d x

\frac{1}{2} (- \frac{1}{9} cos (9 x) - \frac{1}{5} cos (5 x)) + C

Solution steps

\int sin (7 x) cos (2 x) d x

Use the following identity:

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

= \int \frac{sin ( 7 x + 2 x ) + sin ( 7 x - 2 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 (7 x + 2 x) + sin (7 x - 2 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 (7 x + 2 x) d x + \int sin (7 x - 2 x) d x)

\int sin (7 x + 2 x) d x = - \frac{1}{9} cos (9 x)

\int sin (7 x - 2 x) d x = - \frac{1}{5} cos (5 x)

= \frac{1}{2} (- \frac{1}{9} cos (9 x) - \frac{1}{5} cos (5 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{9} cos (9 x) - \frac{1}{5} cos (5 x)) + C

d er i v a t i v e 3 t^{2} + 2 t^{3}

\int_{0}^{1} \frac{2 x}{( x ^{2} + 2 ) ^{2}} d x

\frac{\partial}{\partial x} (7 e^{5 x})

f (x) = e^{x^{2} - 4}

\frac{\partial}{\partial x} (x^{2} + y^{2} - 4 y)

What is the integral of sin(7x)cos(2x) ?
The integral of sin(7x)cos(2x) is 1/2 (-1/9 cos(9x)-1/5 cos(5x))+C