What is the integral of sin(2x)cos(5x) ?

The integral of sin(2x)cos(5x) is 1/2 (-1/7 cos(7x)+1/3 cos(3x))+C

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integral of sin(2x)cos(5x)

\int sin (2 x) cos (5 x) d x

\frac{1}{2} (- \frac{1}{7} cos (7 x) + \frac{1}{3} cos (3 x)) + C

Solution steps

\int sin (2 x) cos (5 x) d x

Use the following identity:

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

= \int \frac{sin ( 2 x + 5 x ) + sin ( 2 x - 5 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 (2 x + 5 x) + sin (2 x - 5 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 (2 x + 5 x) d x + \int sin (2 x - 5 x) d x)

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

\int sin (2 x - 5 x) d x = \frac{1}{3} cos (3 x)

= \frac{1}{2} (- \frac{1}{7} cos (7 x) + \frac{1}{3} cos (3 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{7} cos (7 x) + \frac{1}{3} cos (3 x)) + C

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

\frac{d}{d x} (4 x \cdot sin (x))

\int - sin (3 x) d x

n = 0 \sum \infty \frac{t ^{n}}{n !}

\frac{d}{d x} ((x + 1 + x - 1)^{4})

What is the integral of sin(2x)cos(5x) ?
The integral of sin(2x)cos(5x) is 1/2 (-1/7 cos(7x)+1/3 cos(3x))+C