# Determine `int cos(3x+4) dx` using substitution.

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The integral `int cos(3x+4) dx` has to be determined.

It is known that the integral of cos x is sin x. In cos (3x+4) the cosine of a function of x is being considered.

Use substitution here; let y = 3x + 4

`dy = 3 dx`

`int cos(3x+4) dx`

= `int cos y * (1/3)* dy`

= `(1/3)*int cos y dy`

= `(1/3)*sin y`

replace y by 3x + 4

= `(1/3)*sin(3x+4)`

**The integral `int cos(3x+4) dx` is `(sin(3x+4))/3 + C` **

Before we start, we should know that the derivative of cos u is sin u+C.

- Use u-substitution

- u=3x+4
- Find derivative: du=3dx
- `(1/3)du=dx` &simplify.

- Rewrite equ with u

- `(1/3)intcosudu`
- Integrate: (1/3) sinu+C``

- Plug-in what we had as u

- (1/3) sin(3x+4)+C

- ` `

To use u substitution, you set u = 3x + 4, then take the derivative of u, which is du = 3dx, which means dx = du/3. Plug these back into the original integration.

(1/3) `int` cos(u)du

Then solve:

(1/3)sin(u) + C

Plug what u is back in:

(1/3)sin(3x + 4) + C