# Homework Help: How about this U sub

1. Sep 12, 2007

### Saladsamurai

$$\int sin^32\theta d\theta$$

I am probably overlooking the "easy" way...but as of now I am looking at double angle rules....power reduction.....
I know it can't be as hard as I am making it.

Casey

2. Sep 12, 2007

### l46kok

I really don't think you can do this problem without using trig identities.

3. Sep 12, 2007

### Saladsamurai

yeah I was looking at $$sin2\theta=\frac{2tan\theta}{1+tan^2\theta}$$

so $$\int (\frac{2tan\theta}{1+tan^2\theta})^3$$ which looks kind of like an inverse trig integral...but I don't know how to deal with the power of 3.....

4. Sep 12, 2007

### Saladsamurai

That looks easier....u=sin(theta)?....I still don't know how to deal with the
^3

5. Sep 12, 2007

### l46kok

yeah I realized that too and deleted my post. lol

I think power reduction would be the easiest way to go.
$$\sin^3(2\theta) = \frac{3\sin(2\theta)-\sin(6\theta)}{4}$$

You can integrate individually and not have to deal with u-substitution at all.

Last edited: Sep 12, 2007
6. Sep 12, 2007

### Saladsamurai

I am pretty sure you cannot integrate this^^^ individually...but I just realized something retarded
$$\int sin^32\theta d\theta\\ = \int sin^22\theta sin2\theta\\ = \int(1-cos^22\theta)(sin2\theta)$$

I've still got to figure out the u-sub...but I think this should work out.

7. Sep 12, 2007

### l46kok

? Yes you can. You can integrate what's given above.

But your solution is valid also. Just set $$u = cos(2\theta)$$, then you can solve it.

Earlier, I gave you a solution for double angle formula, it turns out that it works too (But kind of redundant)

$$\sin^32(\theta) = (2\sin(\theta)\cos(\theta))^3 = 8sin^3(\theta)cos^3(\theta)$$

$$8sin^3(\theta)cos^3(\theta) = 8sin^3(\theta)cos(\theta)*cos^2(\theta) = 8sin^3(\theta)cos(\theta) * (1-sin^2(\theta))$$

$$=8sin^3(\theta)cos(\theta) - 8sin^5(\theta)cos(\theta)$$

Then, set $$u = sin(\theta)$$

8. Sep 12, 2007

### Saladsamurai

146kok: What I meant is, if there is a way to integrate that individually, I have not learned it yet. The only things I have learned are about 14 or 15 very specific formulas that the integral has to made to match "perfectly".

Thanks for the heads up.
Casey

9. Sep 12, 2007

### l46kok

Oh you probably know how. This is what I meant

$$sin^3(2\theta) = \int\frac{3\sin(2\theta)-\sin(6\theta)}{4}$$

$$\int\frac{3\sin(2\theta)-\sin(6\theta)}{4}\ = \frac{1}{4}(\int3\sin(2\theta)-\int\sin(6\theta)})$$

Can you integrate
$$\int3\sin(2\theta)$$

and

$$\int\sin(6\theta)$$
?

10. Sep 12, 2007

### Kurdt

Staff Emeritus
The power reduction l46kok suggested is the best way to go. Then if you wanted you could do individual subs for 6theta and 2theta.

11. Sep 12, 2007

### Saladsamurai

I have only done single substitions, like the others I have been posting. Like this https://www.physicsforums.com/showthread.php?t=183925

How would you do ^^^ do you use u for all of them, or choose different variables for each sub...?

I wouldn't mind learning this method if you care to walk me through.

Casey

Last edited: Sep 12, 2007
12. Sep 12, 2007

### Kurdt

Staff Emeritus
Well one of the properties of integrals is the following:

$$\int(f(x)+g(x)) dx = \int f(x) dx + \int g(x) dx$$

So you can treat them both as separate integrals and do two separate substitutions. Once you've done a substitution on these functions where they have multiples of their arguments you will rarely need to do so in the future as they are relatively simple. It is of course a good exercise to see where they come from.

Last edited: Sep 12, 2007
13. Sep 12, 2007

### Saladsamurai

$$\int 3sin2\thetad\theta=-\frac{3}{2}cos2\theta+c$$

and

$$\int sin6\thetad\theta=-\frac{1}{6}cos6\theta$$

Ummm....

14. Sep 12, 2007

### Saladsamurai

Oh yeah, I see where this is going...so does the 1/4 get distributed to both of these?...and both "c"s are combined as one?

so final answer is
$$-\frac{3}{8}cos2\theta+\frac{1}{24}cos6\theta+C$$

15. Sep 12, 2007

### Kurdt

Staff Emeritus
That is correct. So remembering both are multiplied by a quarter, what is the final answer?

EDIT: Ok you beat me to it and you have the correct answer. And yes both constants will be combined, since they are unknown numbers it doesn't require us to unnecessarily write them as a sum of two numbers.

Last edited: Sep 12, 2007
16. Sep 12, 2007

### Saladsamurai

$$-\frac{3}{8}cos2\theta+\frac{1}{24}cos6\theta+C$$

guess not......the text answer is not even close......let me check it over

17. Sep 12, 2007

### Saladsamurai

The text solution is not even close. I do not think you can use the power reduction formulas if there is a double angle as the argument. I think you would have to sub in a double angle formula into the power formula.
The text answer is $$-\frac{1}{2}cos2\theta+\frac{1}{6}cos^32\theta+C$$

18. Sep 12, 2007

### Kurdt

Staff Emeritus
What answer does the book give? They are sometimes published with mistakes.

19. Sep 12, 2007

### Saladsamurai

post 17

20. Sep 12, 2007

### Kurdt

Staff Emeritus
OK I've just done it the other way that was suggested and got the answer in the book. I'm still stumped why it doesn't work the other way. I'll go through it again.

21. Sep 12, 2007

### l46kok

You can use power reduction formula regardless of what angle is given.

They are the same answer. Use trig identity on the answer given from back of the book and you will see that they come out to be the same (Provided you integrated correctly)

22. Sep 12, 2007

### Saladsamurai

hmmm.,...what if we Power reduced the textbook solution $$-\frac{1}{2}cos2\theta+\frac{1}{6}cos^32\theta+C$$

will that be $$-\frac{3}{8}cos2\theta+\frac{1}{24}cos6\theta+C$$?

23. Sep 12, 2007

### l46kok

Yes and that is what I told you, if you integrated correctly.

Usually, when you do integration with trigs in it, you will see many different forms of answers, yet in essence, they are the same.

24. Sep 12, 2007

### Kurdt

Staff Emeritus
I must be out of practise I didn't even consider trig subs

Crisis over.

25. Sep 12, 2007

### l46kok

Aww we all have our bad days don't be so down

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