Integration by Parts with a Touch of Trig

[NdrSiege]

Hey y'all. I'm new to the forum, and have a problem that I've been working on all night long. I'm having issues previewing the Latex, so bear with me. I'll post the work I've done so far if the problem code shows up. Thanks.

\int{\sin^{\frac{3}{2}}2\theta\cos^{3}2\theta} d\theta

Now, the fraction power in the sin is really throwing me off. I've tried rewriting the problem like this:

\int{\sqrt{\sin^{3}2\theta}\cos^{3}2\theta} d\theta

Singled out the sine in the square root:

\int{\sqrt{\sin2\theta\sin^{2}2\theta}\cos^{3}2\theta} d\theta

Square rooted the squared sine function:

\int{\sqrt{\sin2\theta}\sin2\theta\cos^{3}2\theta} d\theta

Skipping some steps (summarized): I split the cosine into two parts (\cos^{2}2\theta \cos2\theta) and then used the trig identity to change \cos^{2}2\theta to (1-\sin^{2}2\theta). Put everything together and got this:

\int{\sqrt{\sin2\theta}\sin2\theta\cos2\theta-sin^{3}2\theta\cos2\theta d\theta}

And now I'm completely lost. I'm not sure if I'm moving in the right direction or just making it a more difficult problem then it really is.

Now that I think about it, I'm not sure if I use integration by parts here.

 

[Kreizhn]

The easiest way I can think about doing this integral requires a bit of a trick. Have you ever seen the proof as to why

\int e^{x^2} dx =\sqrt{2\pi} ?

We will use the same trick here. As a matter of fact, you yourself have said that you don't like the fractional power, so let's get rid of it.

Let f(x)=\int{\sin^{\frac{3}{2}}2\theta\cos^{3}2\theta} d\theta

Then [f(x)]^2=\int{\sin^3 2\theta \cos^6 2\theta} d\theta

We'll save one sine, and then use the identity

sin^2 2\theta = 1-cos^2 2\theta

this gives

[f(x)]^2=\int{\sin 2\theta \sin^2 2\theta \cos^6 2\theta} d\theta

[f(x)]^2=\int{\sin 2\theta (1-cos^2 2\theta) \cos^6 2\theta} d\theta

[f(x)]^2=\int{\sin 2\theta ( \cos^6 2 \theta-cos^8 2\theta) d\theta

now make the substitution u = cos 2\theta and everything follows easily from there. Once you have your solution, simply substitute back to trigonometric functions, and take the square root to return to f(x).

 

[NdrSiege]

Thanks for the help. I ran into another brick wall though. I worked the problem down to the part where I get this:

[f(x)]^2=\int{\sin 2\theta ( \cos^6 2 \theta-cos^8 2\theta) d\theta

And substitute:

u: u = \cos 2\theta

du: du = -2\sin 2\theta

du: -\frac{1}{2}du = \sin 2\theta

So I'm left with the following integral:

-\frac{1}{2} \int { u^6 - u^8 } du

Which I integrate and get the following:

-\frac{1}{2} ( \frac{1}{7} u^7 - \frac{1}{9} u^9 ) \\ <br /> =-\frac{1}{14} u^7 + \frac{1}{18} u^9

I then squareroot the entire function, because we squared is earlier. And I'm left with:

-\frac{\sqrt{14}}{14} cos^\frac{7}{2}2\theta + \frac{\sqrt{2}}{6} cos^\frac{9}{2}2\theta + C

This looks really ugly, and when I plug it into a graph, it doesn't match up. I even tried plugging it in and simplifying it in my CAS and still no luck. I'm not asking for the answer, I just want to know if I goofed up somewhere.

 

[NdrSiege]

It looks like I goofed up the Latex too... I can't seem to fix it.

 

[cristo]

Is this what you meant? (You missed a bracket after \sqrt{2})

-\frac{\sqrt{14}}{14} cos^\frac{7}{2}2\theta + \frac{\sqrt{2}}{6} cos^\frac{9}{2}2\theta + C

 

[NdrSiege]

Yea, that's it. My latex image won't refresh.

 

[Kreizhn]

Remember, \sqrt{x+y} \neq \sqrt{x}+\sqrt{y}, in general

 

[HallsofIvy]

The easiest way I can think about doing this integral requires a bit of a trick. Have you ever seen the proof as to why

\int e^{x^2} dx =\sqrt{2\pi} ?

You meant, of course,
\int_0^\infty e^{-x^2}dx= \sqrt{2\pi}

Let f(x)=\int{\sin^{\frac{3}{2}}2\theta\cos^{3}2\theta } d\theta
Then [f(x)]^2=\int{\sin^3 2\theta \cos^6 2\theta} d\theta

Are you sure that's true? That is NOT the "trick" that is used in the e^{-x^2} integration you referred to.

If f(x)= \int x dx= \frac{1}{2}x^2, then [f(x)]^2= \frac{1}{4}x^4 which is NOT \int x^2 dx!

 

[Kreizhn]

Thank you for the correction to the e^(-x^2); however it is indeed the trick used to solve that integral, though I admit that I did not apply it properly in this case.

The trick is indeed to square the function, giving you a product of two integrals of independent variables x and y, then to make a change of co-ordinates to a polar representation so that the Jacobian gives us an elementary anti-derivative.

For your example

f(x)= \int x dx= \frac{1}{2}x^2[/tex]<br /> [f(x)]^2 = \displaystyle \int \int xy \,dx dy<br /> =\int \frac{1}{2}x^2 y dy<br /> =\frac{1}{4}x^2y^2<br /> \Rightarrow f(x) = \frac{1}{2} x*y<br /> <br /> Which, when x=y is indeed the correct solution.