What is the integral of sin(8x)cos(2x) ?

The integral of sin(8x)cos(2x) is 1/2 (-1/10 cos(10x)-1/6 cos(6x))+C

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integral of sin(8x)cos(2x)

\int sin (8 x) cos (2 x) d x

\frac{1}{2} (- \frac{1}{10} cos (10 x) - \frac{1}{6} cos (6 x)) + C

Solution steps

\int sin (8 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 ( 8 x + 2 x ) + sin ( 8 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 (8 x + 2 x) + sin (8 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 (8 x + 2 x) d x + \int sin (8 x - 2 x) d x)

\int sin (8 x + 2 x) d x = - \frac{1}{10} cos (10 x)

\int sin (8 x - 2 x) d x = - \frac{1}{6} cos (6 x)

= \frac{1}{2} (- \frac{1}{10} cos (10 x) - \frac{1}{6} cos (6 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{10} cos (10 x) - \frac{1}{6} cos (6 x)) + C

x^{^{''}} + 5 x = 0

x^{2} sin (x) tan (x)

y = \frac{1}{27} (9 x^{2} + 6)^{\frac{3}{2}}

\frac{d}{d x} (tan (\frac{2}{x}))

\int \frac{x - 6}{x ^{2} - 16 x + 65} d x

What is the integral of sin(8x)cos(2x) ?
The integral of sin(8x)cos(2x) is 1/2 (-1/10 cos(10x)-1/6 cos(6x))+C