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

Question

5407 views around the world

You can reuse this answer
Creative Commons License

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