Integrating Trigonometric Functions with Double Angle and Power Reduction Rules

[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

 

[asd1249jf]

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

 

[Saladsamurai]

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

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...

 

[Saladsamurai]

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

 

[asd1249jf]

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.

 

[Saladsamurai]

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.

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.

 

[asd1249jf]

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.

? 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 $$u = sin(\theta)$$

 

[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

 

[asd1249jf]

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

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

 

[Kurdt]

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

 

[Saladsamurai]

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

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

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

 

[Kurdt]

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.

 

[Saladsamurai]

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

and

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

Ummm...

 

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

 

[Kurdt]

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.

 

[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

 

[Saladsamurai]

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.

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 $$-\frac{1}{2}cos2\theta+\frac{1}{6}cos^32\theta+C$$

 

[Kurdt]

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


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

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

 

[Saladsamurai]

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

post 17

 

[Kurdt]

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.

 

[asd1249jf]

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 $$-\frac{1}{2}cos2\theta+\frac{1}{6}cos^32\theta+C$$

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)

 

[Saladsamurai]

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)

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

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

 

[asd1249jf]

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

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

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.

 

[Kurdt]

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

Crisis over.

 

[asd1249jf]

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

Crisis over.

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

 

[Saladsamurai]

Thanks for all the help guys. Though I still think this is the simplest mode of attack


$$\int sin^32\theta d\theta\\ = \int sin^22\theta sin2\thetad\theta\\ = \int(1-cos^22\theta)sin2\theta d\theta$$
where $$u=cos2\theta$$

But everybody has their preferences.

Thanks again,
Casey

 

[rocomath]

i just tried it that way, I'm not getting an answer, lol

 

[Kurdt]

i just tried it that way, I'm not getting an answer, lol

It does work. I'll take you through it here.

$$\int sin^3(2\theta) d\theta$$

If we set $u = 2\theta$, therefore:

$$du=2 d\theta$$

$$\frac{1}{2} \int sin^3(u) du = \frac{1}{2} \int sin^2(u)sin(u) du = \frac{1}{2} \int (1-cos^2(u))sin(u) du$$

Now using the substitution $v = cos(u)$

$$dv= -sin(u) du$$

$$-\frac{1}{2}\int (1-v^2) dv = -\frac{1}{2} \left( v- \frac{v^3}{3} \right) +C$$

And finally:

$$\frac{cos^3(2\theta)}{6}-\frac{cos(2\theta)}{2} + C$$

 

[rocomath]

amazing. thanks for the insight, i haven't used 2 subst. for an integral.