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

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Answer 1 · 2017-03-01T02:48:35

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

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

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