What is the integral of-284244sin(120pit+pi/3) ?

The integral of-284244sin(120pit+pi/3) is -284244(1/(80sqrt(3)pi)sin(120pit)-1/(240pi)cos(120pit))+C

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integral of-284244sin(120pit+pi/3)

\int - 284244 sin (120 π t + \frac{π}{3}) d t

- 284244 (\frac{1}{80 3 π} sin (120 π t) - \frac{1}{240 π} cos (120 π t)) + C

Solution steps

\int - 284244 sin (120 π t + \frac{π}{3}) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 284244 \cdot \int sin (120 π t + \frac{π}{3}) d t

Rewrite using trig identities

= - 284244 \cdot \int \frac{3}{2} cos (120 π t) + \frac{1}{2} sin (120 π 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

= - 284244 (\int \frac{3}{2} cos (120 π t) d t + \int \frac{1}{2} sin (120 π t) d t)

\int \frac{3}{2} cos (120 π t) d t = \frac{1}{80 3 π} sin (120 π t)

\int \frac{1}{2} sin (120 π t) d t = - \frac{1}{240 π} cos (120 π t)

= - 284244 (\frac{1}{80 3 π} sin (120 π t) - \frac{1}{240 π} cos (120 π t))

Add a constant to the solution

= - 284244 (\frac{1}{80 3 π} sin (120 π t) - \frac{1}{240 π} cos (120 π t)) + C

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What is the integral of-284244sin(120pit+pi/3) ?
The integral of-284244sin(120pit+pi/3) is -284244(1/(80sqrt(3)pi)sin(120pit)-1/(240pi)cos(120pit))+C