Simplifying Trigonometric Integrals Using Trig Identities

In summary, the conversation is discussing two integration problems and various approaches to solving them. The first problem involves using a trigonometric identity and u-substitution, while the second problem involves using a trigonometric identity and a substitution of u=sin(3x). There is also a discussion about a possible typo in the solution manual for the first problem. In the end, the person was able to solve the problems successfully.
  • #1
Saladsamurai
3,020
7
Okay so I am suposed to evaluate 2 of these:

1.) [tex]\int\sin^3(a\theta)d\theta[/tex] the solution manual looks

like it used a trig ID to do this. Why??

Doesn't [tex]\int\sin^3u du=\frac{1}{3}\cos^3u-(\cos u)+C[/tex] ?

So just use the u-sub [tex]u=a\theta \Rightarrow du/a=d\theta[/tex]

So it should just be [tex]\frac{1}{3a}\cos^3a\theta-\frac{1}{a}\cosa\theta+C[/tex] ?But they got: [tex]-\frac{1}{a}\cos a\theta-\frac{1}{3a}\cos^3a\theta+C[/tex]

Why the negative -1/3a ? What did I miss? Is it my formula?
2.) For [tex]\int\sin^4(3x)\cos^3xdx[/tex] they went from <---that to

[tex]=\int\sin^4(3x)(1-sin^23x)\cos3xdx[/tex]

How the hell does that make this problem ANY easier?! Is there a formula for ^^^that!Casey
 
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  • #2
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  • #3
You're using sin(u) like a variable, and integrating, which will certainly get you nowhere. One thing is that [tex] \int sin(u)^{3}[/tex] does not equal [tex]\frac{1}{3}\cos^3u[/tex]. You're going to have to use u-substitution, as you already have, and almost certainly some integration by parts.
 
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  • #4
I would suggest you use the trig identity sin(x)^2=1-cos(x)^2, followed by u=cos(x).
 
  • #5
[tex]=\int\sin^4(3x)(1-sin^23x)\cos3xdx[/tex]If you let u=sin3x does that make it any easier?
 
  • #6
Dick said:
I would suggest you use the trig identity sin(x)^2=1-cos(x)^2, followed by u=cos(x).

I'll try that now. But why is my formula not producing the correct result if I am using u=a*theta ? Shouldn't it be 1/a *{Pug it in} ?
 
  • #7
[tex]\int \sin^3(a\theta)d\theta = \int \sin(a\theta)d\theta - \int \sin(a\theta)\cos^2(a\theta)d\theta[/tex]

The first term should be easy for you to integrate, for the second term, try a substitution of

[tex]u = \cos(a\theta) \Rightarrow \frac{du}{d\theta} = -a\sin(a\theta) \Rightarrow \sin(a\theta)d\theta = -\frac{1}{a}du[/tex]

After which the equation simplifies a little.

Use a similar idea for the second one.
 
  • #8
Dick said:
I would suggest you use the trig identity sin(x)^2=1-cos(x)^2, followed by u=cos(x).

So I got it using this. But I am still trying to figure out where I went wrong with the formula.

Casey
 
  • #9
Saladsamurai said:
So I got it using this. But I am still trying to figure out where I went wrong with the formula.

Casey

Seems to me like there is simply a typo in the book (the coefficient of the cos^3 should indeed be positive)
 
  • #10
nrqed said:
Seems to me like there is simply a typo in the book (the coefficient of the cos^3 should indeed be positive)

Not if you use the trig substitution; it is negative. It is definitely a mistake on my part.

Casey
 
  • #11
No, it's positive. (1-cos^(x)^2)*sin(x)*dx. u=cos(x), du=-sin(x)*dx. The u^3/3 term comes out with a plus sign.
 
  • #12
Saladsamurai said:
2.) For [tex]\int\sin^4(3x)\cos^3xdx[/tex] they went from <---that to

[tex]=\int\sin^4(3x)(1-sin^23x)\cos3xdx[/tex]

did you mean [tex]\cos^{3}3xdx[/tex]?
 
  • #13
rocophysics said:
did you mean [tex]\cos^{3}3xdx[/tex]?

I don't know what I meant?! I figured it out though.

Thanks!
Casey
 
  • #14
i was trying to solve it for myself but then it saw that ... lol, is all good though.
 

What are the basic principles of integrating powers of sin?

The basic principle of integrating powers of sin is to use the trigonometric identity sin2(x) + cos2(x) = 1 to simplify the integral, and then use u-substitution or trigonometric substitution to solve it.

How do I determine the appropriate substitution to use when integrating powers of sin?

The substitution to use depends on the form of the integral. If the integral contains an odd power of sin, u-substitution is typically used. If the integral contains an even power of sin, trigonometric substitution is typically used.

Can I use integration by parts to integrate powers of sin?

No, integration by parts cannot be used to integrate powers of sin directly. However, it can be used in combination with other techniques, such as u-substitution, to solve certain integrals.

Are there any special cases when integrating powers of sin?

Yes, there are two special cases to note: when the power of sin is 0, the integral simplifies to just x + C, and when the power of sin is 1, the integral simplifies to -cos(x) + C.

What are some common mistakes to avoid when integrating powers of sin?

Some common mistakes include forgetting to use the trigonometric identity, using u-substitution when trigonometric substitution is needed, and integrating the power of sin incorrectly. It is important to carefully follow the steps and check your work to avoid mistakes.

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