What is the integral of sin(7x)cos(5x) ?

The integral of sin(7x)cos(5x) is 1/2 (-1/12 cos(12x)-1/2 cos(2x))+C

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integral of sin(7x)cos(5x)

\int sin (7 x) cos (5 x) d x

\frac{1}{2} (- \frac{1}{12} cos (12 x) - \frac{1}{2} cos (2 x)) + C

Solution steps

\int sin (7 x) cos (5 x) d x

Use the following identity:

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

= \int \frac{sin ( 7 x + 5 x ) + sin ( 7 x - 5 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 (7 x + 5 x) + sin (7 x - 5 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (12 x) + sin (2 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 (12 x) d x + \int sin (2 x) d x)

\int sin (12 x) d x = - \frac{1}{12} cos (12 x)

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

= \frac{1}{2} (- \frac{1}{12} cos (12 x) - \frac{1}{2} cos (2 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{12} cos (12 x) - \frac{1}{2} cos (2 x)) + C

h \to 0 lim (\frac{3 e ^{h} - 3}{h})

\int_{- \infty}^{0} \frac{x}{x ^{4} + 4} d x

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

\int_{l n (3)}^{l n (5)} e^{2 x} d x

f (x) = 3^{x}, at x = 0

What is the integral of sin(7x)cos(5x) ?
The integral of sin(7x)cos(5x) is 1/2 (-1/12 cos(12x)-1/2 cos(2x))+C