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

The integral of cos(x)sin(kx) is 1/2 (-1/(k+1)cos(kx+x)-1/(k-1)cos(kx-x))+C

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integral of cos(x)sin(kx)

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

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

Solution steps

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

Use the following identity:

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

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

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

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

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

Add a constant to the solution

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

0.0588 s^{1.125}

\int_{0}^{\frac{1}{2}} 4 cos (π x) d x

\int x \cdot \frac{1}{x ^{2}} d x

x \to - 3 lim (- 2 x^{3})

f (x) = x + 2, at x = - 1

What is the integral of cos(x)sin(kx) ?
The integral of cos(x)sin(kx) is 1/2 (-1/(k+1)cos(kx+x)-1/(k-1)cos(kx-x))+C