What is the integral of sin(7x)cos(4x) ?

The integral of sin(7x)cos(4x) is 1/2 (-1/11 cos(11x)-1/3 cos(3x))+C

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integral of sin(7x)cos(4x)

\int sin (7 x) cos (4 x) d x

\frac{1}{2} (- \frac{1}{11} cos (11 x) - \frac{1}{3} cos (3 x)) + C

Solution steps

\int sin (7 x) cos (4 x) d x

Use the following identity:

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

= \int \frac{sin ( 7 x + 4 x ) + sin ( 7 x - 4 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 + 4 x) + sin (7 x - 4 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (11 x) + sin (3 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 (11 x) d x + \int sin (3 x) d x)

\int sin (11 x) d x = - \frac{1}{11} cos (11 x)

\int sin (3 x) d x = - \frac{1}{3} cos (3 x)

= \frac{1}{2} (- \frac{1}{11} cos (11 x) - \frac{1}{3} cos (3 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{11} cos (11 x) - \frac{1}{3} cos (3 x)) + C

\int x^{3} e^{- 3 x} d x

y^{^{''}} + y = csc^{2} (t)

(1.4) (5.7)

\int_{- 1}^{1} (x^{2} - 2) d x

x \to 0 lim (\frac{sin ( x ^{4} )}{x})

What is the integral of sin(7x)cos(4x) ?
The integral of sin(7x)cos(4x) is 1/2 (-1/11 cos(11x)-1/3 cos(3x))+C