What is the integral of sin(-4x)cos(3x) ?

The integral of sin(-4x)cos(3x) is 1/2 (1/7 cos(7x)+cos(x))+C

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integral of sin(-4x)cos(3x)

\int sin (- 4 x) cos (3 x) d x

\frac{1}{2} (\frac{1}{7} cos (7 x) + cos (x)) + C

Solution steps

\int sin (- 4 x) cos (3 x) d x

Use the following identity:

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

= \int \frac{sin ( - 4 x + 3 x ) + sin ( - 4 x - 3 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 (- 4 x + 3 x) + sin (- 4 x - 3 x) d x

Simplify

sin (- 4 x + 3 x) + sin (- 4 x - 3 x) : - sin (7 x) - sin (x)

= \frac{1}{2} \cdot \int - sin (7 x) - sin (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 (7 x) d x - \int sin (x) d x)

\int sin (7 x) d x = - \frac{1}{7} cos (7 x)

\int sin (x) d x = - cos (x)

= \frac{1}{2} (- (- \frac{1}{7} cos (7 x)) - (- cos (x)))

Simplify

= \frac{1}{2} (\frac{1}{7} cos (7 x) + cos (x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{7} cos (7 x) + cos (x)) + C

\frac{\partial}{\partial y} (e^{x} (x - y)^{2})

\int (cos (x) + 3^{x}) d x

x \to 0 lim (\frac{( 2 - x ) ^{2} + 1}{x})

\frac{d}{d x} (arcsec (7 ln (4 x)))

\int_{- \infty}^{\infty} 15 x e^{- x^{2}} d x

What is the integral of sin(-4x)cos(3x) ?
The integral of sin(-4x)cos(3x) is 1/2 (1/7 cos(7x)+cos(x))+C