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

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

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integral of sin(x)cos(kx)

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

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

Solution steps

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

Use the following identity:

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

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

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

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

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

Add a constant to the solution

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

\int 9 (9 x + 4) d x

f (x) = x^{2} - 2 cos (x) - 3

x y^{2} y^{^{'}} = y^{3} - x^{3}

\int \frac{( 1 )}{25 x ^{2} - 16} d x

\int \frac{4 x}{2 x ^{2} + 1} d x

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