What is the integral of-10cos(4t+pi/6) ?

The integral of-10cos(4t+pi/6) is -5/2 sin(4t+pi/6)+C

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integral of-10cos(4t+pi/6)

\int - 10 cos (4 t + \frac{π}{6}) d t

- \frac{5}{2} sin (4 t + \frac{π}{6}) + C

Solution steps

\int - 10 cos (4 t + \frac{π}{6}) d t

Take the constant out:

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

= - 10 \cdot \int cos (4 t + \frac{π}{6}) d t

Apply u-substitution

: \int cos (u) \frac{1}{4} d u

= - 10 \cdot \int cos (u) \frac{1}{4} d u

Take the constant out:

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

= - 10 \cdot \frac{1}{4} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= - 10 \cdot \frac{1}{4} sin (u)

Substitute back

u = 4 t + \frac{π}{6}

= - 10 \cdot \frac{1}{4} sin (4 t + \frac{π}{6})

Simplify

- 10 \cdot \frac{1}{4} sin (4 t + \frac{π}{6}) : - \frac{5}{2} sin (4 t + \frac{π}{6})

= - \frac{5}{2} sin (4 t + \frac{π}{6})

Add a constant to the solution

= - \frac{5}{2} sin (4 t + \frac{π}{6}) + C

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What is the integral of-10cos(4t+pi/6) ?
The integral of-10cos(4t+pi/6) is -5/2 sin(4t+pi/6)+C