What is the integral of sin(4t)cos(5t) ?

The integral of sin(4t)cos(5t) is 1/2 (-1/9 cos(9t)+cos(t))+C

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integral of sin(4t)cos(5t)

\int sin (4 t) cos (5 t) d t

\frac{1}{2} (- \frac{1}{9} cos (9 t) + cos (t)) + C

Solution steps

\int sin (4 t) cos (5 t) d t

Use the following identity:

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

= \int \frac{sin ( 4 t + 5 t ) + sin ( 4 t - 5 t )}{2} d t

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 t + 5 t) + sin (4 t - 5 t) d t

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 (4 t + 5 t) d t + \int sin (4 t - 5 t) d t)

\int sin (4 t + 5 t) d t = - \frac{1}{9} cos (9 t)

\int sin (4 t - 5 t) d t = cos (t)

= \frac{1}{2} (- \frac{1}{9} cos (9 t) + cos (t))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{9} cos (9 t) + cos (t)) + C

\int \frac{x + 18}{x ^{2} + 6 x + 13} d x

n = 1 \sum \infty \frac{1}{( n + 2 )}

\int x (2 - x) d x

\frac{\partial}{\partial x} (\frac{18 x}{( y + 9 z ) ^{3}})

\int_{0}^{\infty} x^{0} e^{- x} d x

What is the integral of sin(4t)cos(5t) ?
The integral of sin(4t)cos(5t) is 1/2 (-1/9 cos(9t)+cos(t))+C