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

The integral of sin(x)cos(5x) is 1/2 (-1/6 cos(6x)+1/4 cos(4x))+C

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integral of sin(x)cos(5x)

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

\frac{1}{2} (- \frac{1}{6} cos (6 x) + \frac{1}{4} cos (4 x)) + C

Solution steps

\int sin (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 ( x + 5 x ) + sin ( 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 (x + 5 x) + sin (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 (x + 5 x) d x + \int sin (x - 5 x) d x)

\int sin (x + 5 x) d x = - \frac{1}{6} cos (6 x)

\int sin (x - 5 x) d x = \frac{1}{4} cos (4 x)

= \frac{1}{2} (- \frac{1}{6} cos (6 x) + \frac{1}{4} cos (4 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{6} cos (6 x) + \frac{1}{4} cos (4 x)) + C

x \to 0 lim (x^{a})

\int \frac{1}{6 x ^{3}} d x

\int 18 + 0.5 x d x

\frac{d}{d t} (cos (a t))

\frac{d}{d x} ((x - 4)^{8} (2 x + 3)^{6})

What is the integral of sin(x)cos(5x) ?
The integral of sin(x)cos(5x) is 1/2 (-1/6 cos(6x)+1/4 cos(4x))+C