Question

see question in comment box !
Integrals....

I'm working on it and will post my answer in the comment box when I'm done.

Answer 1

\[\int\limits \frac{ sinx}{ (4 +cosx)^{6}} (dx)\]

Answer 2

\[ u= (4 +cosx)^{6}\]

and

\[\frac{ du }{ dx} = 6(4+cosx)^{5}(-sinx)\]

Answer 3

Nah, just let \(u = 4+\cos x\)

Answer 4

You usually dont want to include the exponent in your u-substitutions. Usually its always whats inside of the power, square, root, etc. In this case that is true as well, just let u 4 + cosx as mentioned above.

Answer 5

what about the exponent power to the 6 ?

Answer 6

Okay cool! Makes it easier, so is this a rule that one should apply when we see exponents on the outside ?

Answer 7

Your goal is to get rid of everything x and dx and to have only u, du, and constant. Using u = (4+cosx)^6 is a chain rule and adds extra expressions with x that cannot be cancelled out.

Answer 8

So you often want your u-substitutions to be selections that wont be chain rules, since those multiply by inner functions, hence creating more expressions of x.

Answer 9

Okay so if u= 4+cosx
that changes my du/dx too !

Answer 10

du/dx = -sin x

Answer 11

Right, it makes du = -sinxdx, which cancels out the necessary sinx expression without creating extra things that cannot cancel.

Answer 12

okay okay cool
now i'm trying to do the u-substition

Answer 13

does the du go on top ?

Answer 14

You actually have to have du on top in order to properly integrate. I suppose its not explicitly mentioned all the time, but whatever your differential is, du, dx, etc, it must be in the numerator in order to integrate.

Answer 15

as in \[\int\limits \frac{ -sinx }{u ^{6} }\]

Answer 16

is this right ?

Answer 17

Still need your expression for dx.

Answer 18

okay