find the indefinite integral

Question

chrisplusian · Answer

$$\int\limits (\cos3\theta-1)d \theta$$

chrisplusian · Answer

solutions guide shows the first step as

anonymous · Answer

u substitution:
u = 3t - 1
du/dt = 3
dt = du/3

chrisplusian · Answer

$$(1/3)\int\limits \cos3\theta(3) d \theta-\int\limits d \theta$$

chrisplusian · Answer

then$$(1/3)\sin 3\theta-\theta +c$$

chrisplusian · Answer

I am having a hard time following the logic in the first step

anonymous · Answer

what dont you get?

anonymous · Answer

Chris, have you studied U substitutions yet?

chrisplusian · Answer

yes i have and I thougth i was pretty good at them till i got to the logarithmic ones

chrisplusian · Answer

now the part i don't get is

chrisplusian · Answer

I agree that u=3t-1

anonymous · Answer

Well, choosing
$$u = 3\theta - 1$$
makes this one work out pretty well.

anonymous · Answer

Take du/dt, then solve that equation for dt.

Substitute in u and dt in your original integral.

chrisplusian · Answer

so in my solution I thought i did that and came up with

chrisplusian · Answer

$$-(1/3)(lnsin3t-1)+c$$

anonymous · Answer

woah woah woah... logarithms don't come into this.

chrisplusian · Answer

but that did not match thier solution and I am trying to follow how they got there

chrisplusian · Answer

this is in the section of my book that is titled "integration of the natural logarithmic function"

chrisplusian · Answer

could you walk me through it then?

anonymous · Answer

u = 3t-1
du/dt = 3
dt = du/3

Use those to substitute into
$$\int\limits \cos(3t-1)dt$$
gives
$$\int\limits \cos(u)/3du$$
=
$$(1/3) \int\limits cosu du$$

chrisplusian · Answer

I ended up with $$(1/3)\int\limits (cosu) du$$
by definition ( i now see) cos u = sin u + c so even then I would have (1/3) sin 3x-1 + c

chrisplusian · Answer

which agrees with what you just put in. the only problem is that it doesn't match the answer they are giving

anonymous · Answer

Right.

anonymous · Answer

Hmm and the answer they give is?

chrisplusian · Answer

there answer:
$$(1/3)\sin3 \theta -\theta +c$$

chrisplusian · Answer

I meant thier answer

chrisplusian · Answer

and yes it is an odd numbered question and I have the solutions guide

anonymous · Answer

Oooh. Simple mistake. The original integral isn't of
cos(3t-1)
it's of
cos(3t)-1

anonymous · Answer

See?

chrisplusian · Answer

the original problem is exactly like this

chrisplusian · Answer

$$\int\limits (\cos3\theta-1)d \theta$$

anonymous · Answer

Yeah. Precisely. That's not cos(3t-1). It's cos(3t)-1

anonymous · Answer

Split that up into two integrals:
$$\int\limits \cos(3t)dt - \int\limits 1 dt$$

chrisplusian · Answer

ah I see says the blind man!!! thanks

anonymous · Answer

Finally, you get it, just split them out:
Int ( f (t) - g (t)dt ) = Int ( f (t) dt ) - Int ( g (t) dt )