Question

Indefinite Integral Help please

Answer 1

\[\int\limits_{}^{}7\sin \frac{ \theta }{ 3 }d\]

Answer 2

d theta at the end sorry

Answer 3

without using U substitution.. where does the theta go? i get -21/theta cos theta/3 +c ... the book has just -21 cos theta/3 + c

Answer 4

\[\frac{ -21 }{ \theta }\cos \frac{ \theta }{ 3 }+C\]

Answer 5

\[-21 \cos \frac{ \theta }{ 3 }+C\] is the solutions manual answer

Answer 6

So you want to know how to do this `without` using a U-sub?

Answer 7

yes, sorry last calc 1 class... so she didnt introduce u sub. just the idea of indef integral and anti derivative

Answer 8

I cant figure out a way to get rid of the denominator.

Answer 9

\[\large \int\limits7\sin\left(\frac{\theta}{3}\right)\;d\theta\]
So we'll use this idea called `advanced guessing`.
The integral will give us `something` like this,\[\large -7\cos\left(\frac{\theta}{3}\right)+C\]

Right?
But there is something going on in the inside that we have to account for.

Answer 10

so we are going to assume theta is 1?

Answer 11

So let's look at our solution a moment.
If we were to take it's derivative, what would happen?

Well let's just try, and find out.
And from there, maybe we can realize a little trick that's happening.

\[\large \frac{d}{dx}\left[-7\cos\left(\frac{\theta}{3}\right)+C\right] \qquad = \qquad 7\sin\left(\frac{\theta}{3}\right)\color{royalblue}{\left(\frac{\theta}{3}\right)'}\]
See that blue term? That comes out as a result of the chain rule.

Answer 12

\[\large =7\sin\left(\frac{\theta}{3}\right)\color{royalblue}{\left(\frac{1}{3}\right)}\]

Answer 13

So when we tried to check our work, an extra 1/3 popped out. We found that our solution we came up with was off by a factor of 3.

To fix this, we'll multiply our solution by 3.

Answer 14

Ok ok all of that aside.. the short answer is, the chain rule in reverse will have you `dividing` by the coefficient instead of multiplying by it.

Answer 15

So instead of multiplying by a factor of 1/3, like the chain rule would normally produce, we instead divide by 1/3.

Answer 16

so the rule for sin -1/k cos kx + C should not be used?

Answer 17

I'm not sure what that is :O

Answer 18

sorry the antiderivative rule for sin.

Answer 19

Here are a couple examples to help drive the point home.\[\large \int\limits \cos(2x)dx \qquad = \frac{1}{2}\sin(2x)+C\]

\[\large \int\limits e^{2x}dx \qquad = \qquad \frac{1}{2}e^{2x}+C\]

When we take derivatives of these functions, normally an extra factor of 2 would pop out right? The opposite is happening when we integrate. See how we're dividing by that factor of 2 in each case?

Answer 20

Is there a special anti-derivative rule for sine? :O Hmm, maybe I should brush up on that. heh

Answer 21

We're dividing by 2 to `compensate` for the 2 that WILL pop out due to the chain rule.
The 1/2 gets rid of that 2.

Answer 22

yes, that's how I got -21/theta. and I use the formula:

\[b \frac{ 1 }{ n+1 }x^n+1\]

for x^n problems.

Answer 23

I think I understand what you're saying.
Like you're trying to do it like this I guess?
Example:\[\large \int\limits \cos(x)dx \qquad \ne \qquad \sin(x)\frac{x^2}{2}+c\]

Applying the Power Rule for Integrals to the inner x?