What is the integral of sin(9t)sin(4t) ?

The integral of sin(9t)sin(4t) is 1/2 (-1/13 sin(13t)+1/5 sin(5t))+C

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integral of sin(9t)sin(4t)

\int sin (9 t) sin (4 t) d t

\frac{1}{2} (- \frac{1}{13} sin (13 t) + \frac{1}{5} sin (5 t)) + C

Solution steps

\int sin (9 t) sin (4 t) d t

Rewrite using trig identities

= \int \frac{1}{2} (- cos (13 t) + cos (5 t)) 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 - cos (13 t) + cos (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 cos (13 t) d t + \int cos (5 t) d t)

\int cos (13 t) d t = \frac{1}{13} sin (13 t)

\int cos (5 t) d t = \frac{1}{5} sin (5 t)

= \frac{1}{2} (- \frac{1}{13} sin (13 t) + \frac{1}{5} sin (5 t))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{13} sin (13 t) + \frac{1}{5} sin (5 t)) + C

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y y^{^{'}} + y^{\frac{3}{2}} = 1

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\int 5 tan^{2} (x) sec (x) d x

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What is the integral of sin(9t)sin(4t) ?
The integral of sin(9t)sin(4t) is 1/2 (-1/13 sin(13t)+1/5 sin(5t))+C