What is the integral of-4cos(2t+pi) ?

The integral of-4cos(2t+pi) is -2sin(2t+pi)+C

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integral of-4cos(2t+pi)

\int - 4 cos (2 t + π) d t

- 2 sin (2 t + π) + C

Solution steps

\int - 4 cos (2 t + π) d t

Take the constant out:

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

= - 4 \cdot \int cos (2 t + π) d t

Apply u-substitution

: \int cos (u) \frac{1}{2} d u

= - 4 \cdot \int cos (u) \frac{1}{2} d u

Take the constant out:

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

= - 4 \cdot \frac{1}{2} \cdot \int cos (u) d u

Use the common integral:

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

= - 4 \cdot \frac{1}{2} sin (u)

Substitute back

u = 2 t + π

= - 4 \cdot \frac{1}{2} sin (2 t + π)

Simplify

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

= - 2 sin (2 t + π)

Add a constant to the solution

= - 2 sin (2 t + π) + C

\int \frac{4400 x}{4 x ^{2} + 320} d x

\frac{\partial}{\partial x} (8 y^{9} x^{7} - 7 y^{8} x^{6})

y^{^{'}} + 3 x^{2} y = sin (x) e^{- x^{3}}

x \to 3 lim (\frac{8 x + 7}{5 x + 7})

\frac{d}{d x} (\frac{1}{4} x^{4} + \frac{1}{8 x ^{2}})

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