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

The integral of cos(x)sin(nx) is 1/2 (-1/(n+1)cos(nx+x)-1/(n-1)cos(nx-x))+C

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integral of cos(x)sin(nx)

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

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

Solution steps

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

Use the following identity:

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

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

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

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

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

Add a constant to the solution

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

\frac{d}{d x} ((cos (3 x)) \cdot 3)

f (x) = - x

ln (\frac{x}{2}), 2

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

\int (x^{2} - 3 x + 2) d x

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