Question

How do you evaluate the integral `int 3sinx+4cosxdx`?

Answer 1

`int 3sin(x)+4cos(x) dx = 4sin(x)-3cos(x) +C`

Explanation:

This integral is easier than it looks!

The first thing that we need to realize is that we can break the integral up over the addition:

`int 3sin(x)+4cos(x) dx = int3sin(x)dx+int 4cos(x) dx`

Then we can move the constants out of the integrals:

`int3sin(x)dx+int 4 cos(x) dx= 3 int sin(x)dx+4 int cos(x) dx`

Now we can just sub in the antiderivatives for `sin(x)` and `cos(x)`.
You should have these memorized, they come up a lot!

`int sin(x) = -cos(x) +c_1`
`int cos(x) = sin(x) + c_2`

`3 int sin(x)dx+4 int cos(x) dx = 3(-cos(x) + c_1) + 4(sin(x) +c_2)`

so the final answer is:

`int 3sin(x)+4cos(x) dx = 4sin(x)-3cos(x) + C`

because this is an indefinite integral, we stop here and do not need to evaluate further.

note: I combined the two constants of integration, `c_1` and `c_2` into one constant `C` because it's easier to deal with