What is the integral of-3cos(4x) ?

The integral of-3cos(4x) is -3/4 sin(4x)+C

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integral of-3cos(4x)

\int - 3 cos (4 x) d x

- \frac{3}{4} sin (4 x) + C

Solution steps

\int - 3 cos (4 x) d x

Take the constant out:

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

= - 3 \cdot \int cos (4 x) d x

Apply u-substitution

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

= - 3 \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

= - 3 \cdot \frac{1}{4} \cdot \int cos (u) d u

Use the common integral:

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

= - 3 \cdot \frac{1}{4} sin (u)

Substitute back

u = 4 x

= - 3 \cdot \frac{1}{4} sin (4 x)

Simplify

- 3 \cdot \frac{1}{4} sin (4 x) : - \frac{3}{4} sin (4 x)

= - \frac{3}{4} sin (4 x)

Add a constant to the solution

= - \frac{3}{4} sin (4 x) + C

\int_{1}^{2} x e^{8 x^{2}} d x

x y^{^{'}} + 2 y = - e^{x}

y^{^{'}} = - 0.5 y

f (x) = x^{2} (- 3 x^{2} - 2)

\int \frac{1}{x ( x + 3 )} d x