Calculus help!

OK, just started up school again after like 2 years, and I refuse to take Calc 1 again, but I'm rusty as hell. Most of it came back, but I got a little stuck on one easy-ass problem, because I forgot exactly how to work the numbers.

integrate 5sin7x - 7cos5x dx

Now, I know it's just easy u substitution, but I forgot how the sandwiched in numbers work... like, I could handle 5sinx, or 5sinx^2, but wtf is 5sin7x? is that 7 just a number I can take out to make 35sinx? That doesn't sound right at all. Someone tell me.

I haven't done Calculus since high school. So I can't help you. In fact, I shouldn't even be here making this post. Good luck!

-5/7 COS 7X - 7/5 SIN 5X + C

Just straight integration.

Derivative of cos(ax) = -a sin (ax)
Derivative of sin(ax) = a cos (ax)

Calculus is just logarithms, differentiation and integration, right?
There's nothing too difficult to get the hang of if that's so, you just have to learn your derivitives for functions like sin, sec, etc:

sin(ax+b) dx = -1/a cos(ax+b) + C
cos(ax+b) dx = 1/a sin(ax+b) +

Yeah, right now it's all pretty easy and stuff I've done before, but it's been a couple years at least since I took Calculus I.

The answer doesn't help me much, that was in the book. It's more the step in between that I'm fuzzy about.

Like, when you integrate 5sin7x, what is that 7? Why do you end up with -5/7 cos7x? sinx to -cosx is obvious, but why the 5/7?

Because the derivative of one of these functions with an argument that isn't just x (or y or whatever) is the derivative of that argument with respect to the variable you're differentating, times the derative of sin/cos/whatever. To make it more clear ;/

Derivative with respect to x of sin (3+2x) = 2 cos (3+2x)

The 2 in front came from 3+2x, as 2 is the derivative of that argument with respect to x. So to integrate, you just go backwards.

Apply: ksin(ax+b) dx = (-1/a) kcos(ax+b) + C
To: 5sin7x
where k = 5, a = 7, b=0

So: (-1/a) kcos(ax+b) + C
Subsitute: (-1/7) 5cos(7x)
where (-1/7)x 5 = -5/7
So: (-5/7)cos7x

THANK YOU <3 (both of you)

Yeah, ok, wow, that made sense. I forgot about that completely.

Just started classes this week after about 2 years of being a hobo after dropping out, to say that I'm rusty would be an incredible understatement.

Expect more from this thread in the future, when the problems actually become difficult.

The same sort of thing applies for integrating cosx, except it's 1/a (not -1/a), where you divide 1 by the multiplier of x inside the brackets

One problem down, several hundred to go.

Good luck Mzo!