What is the integral of 4cos(2x) ?

The integral of 4cos(2x) is 2sin(2x)+C

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integral of 4cos(2x)

\int 4 cos (2 x) d x

2 sin (2 x) + C

Solution steps

\int 4 cos (2 x) d x

Take the constant out:

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

= 4 \cdot \int cos (2 x) d x

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 x

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

Simplify

4 \cdot \frac{1}{2} sin (2 x) : 2 sin (2 x)

= 2 sin (2 x)

Add a constant to the solution

= 2 sin (2 x) + C

\int (2 x - 3)^{- 1} (x + 1)^{- 1} d x

3 x^{3} + 9 x y^{2} = 5 x^{3}

\frac{d}{d x} (e^{\frac{( x ^{2} )}{2}})

\int (5 x - 2)^{3} d x

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} y^{0}