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) \cdot 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

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

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

\int x^{- 2} + 2 x - 6 x^{2} + cos (x) d x

\int (6 sinh (x) + x^{5}) 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