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

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

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

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

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

Solution steps

\int sin (2 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 ( 2 x + 7 x ) + sin ( 2 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 (2 x + 7 x) + sin (2 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 (2 x + 7 x) d x + \int sin (2 x - 7 x) d x)

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

\int sin (2 x - 7 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

\int 5 tan^{5} (x) sec^{4} (x) d x

\int_{6}^{8} \frac{12}{( x - 6 ) ^{3}} d x

y = 6 x^{2} - 2 x, y = x^{3} - 6 x + 2

n \to + (- \infty) + lim (e^{n})

x \to 0 lim (\frac{tan ^{2} ( 3 x )}{x ^{2} cos ( 2 x )})

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