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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \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{\partial}{\partial x} (3 x cos (x y))

g (x) = x (x^{3} - x)

\int (\frac{d y}{1}) d x

\int \frac{2 x ^{2} + 3 x + 3}{( x + 1 ) ^{3}} d x

\int_{0}^{\frac{π}{2}} cos^{6} (x) 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