Question

integration substitution problem, please help! :)

Answer 1

\[\int\limits_{}^{}\cos(2x)-\sin(x)\]

Answer 2

I know this will somehow involve double angle formula, I just need some help getting started I guess. Thanks :)

Answer 3

\[\int\limits_{}^{}(\cos(2x) - \sin(x))dx = \int\limits_{}^{}\cos(2x)dx - \int\limits_{}^{}\sin(x)dx = \frac{\sin(2x)}{2} - (-\cos(x)) = \frac{\sin(2x)}{2}+ \cos(x)\]

Answer 4

I'm a little confused by that, and it gets cut off halfway through

Answer 5

tell me which part you find confusing

Answer 6

You can directly go with separating the integrals... which is the best way to deal this, otherwise use,
\[\cos2x= 1- 2 (sinx)^2\]

Answer 7

Yeah, that's how I wanted to do it, without separating the integrals. So are you saying that I should set the problem up to look like this: \[\int\limits_{}^{}1-2\sin^2(x)-\sin(x)\] ?

Answer 8

whoops forgot the dx at the end

Answer 9

You could separate it into two integrals. That would be much easier:
\[\int\limits \cos(2x)dx - \int\limits \sin(x)\]
for the first one, use u-substitution where u = 2x du = 2
\[\frac{ 1 }{ 2 } \int\limits \cos(u) du - \int\limits \sin(x)dx = \frac{ 1 }{ 2 } \int\limits \cos(u) du - \cos(x) + C\]
you can do the integral of cos yourself. just remember to plug in the original u = 2x.

Answer 10

okay. I'm just curious though, how would I solve it using the equation I set up?

Answer 11

with the double angle substitution?

Answer 12

You don't need to use double angle formula for this.

Answer 13

So I shouldn't do what saloniiigupta95 suggested by replacing cos(2x) with 2-2sin^2x?

Answer 14

whoops I meant 1-2sin^2(x)

Answer 15

You can, but I think that method makes this problem a bit more complex.

Answer 16

Yeah, I just have a feeling that's how my calc professor wants me to solve it though

Answer 17

Ya, you can do it but it is not a good way ...as I mentioned before, first approach is the best...

Answer 18

Either way is fine, strott, I just prefer different methods. I am not sure whethere this is for college or high-school, I usually suggest a shorter method to problems. Since in college exams are usually timed, mine were 50 minutes, and finding a shorter method to a problem allows me time to work on possibly more difficult problems on an exam. But either method is fine. It's which ever makes you more comfortable.

Answer 19

okay, thanks for the help!