Integrate x cos^5(x^2) dx

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hank

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I tried to integrate this by parts, but I get stuck.

I set u = x, du = dx, dv - cos^5(x^2) which leaves me with v = integral of cos^5(x^2). I

This is where I get stuck. Do I do another integration by parts, or do I need to do something else?

Seems like a simple u-sub of u = x^2 won't work here either.

Could there be a reduction formula involved in here?

Maybe do a u-sub to get rid of the x dx which leaves you with int cos^5(u) du and then use the reduction formula on that?
 
had to dust off the CRC for this one ...

start with a u-sub for the x<sup>2</sup>, then use the reduction formula ...

\(\displaystyle \L \int \cos^n{u}du = \frac{1}{n}\cos^{n-1}{u}\sin{u} + \frac{n-1}{n}\int \cos^{n-2}{u}du\)
 
Is there a way to do it without the reduction formula?

Here's what I get...

u = x^2
du = 2x dx
1/2 du = x dx

=1/2 int cos^5u du

= 1/2 [1/5 cos^4usinu + 2/5 int cos^3 u du

Then repeat the process for the int cos^3u du and then go back and fill in the x^2 for the u. Does that sound right?

I come up with this answer:

1/10 cos^4(x^2)sin(x^2) + 2/15 cos^2(x^2)sin(x^2) + 2/15 (x^2) + 4/15 sin2(x^2) + C

According to the book, the answer is: 1/2 sin(x^2) - 1/3 sin^3(x^2) + 1/10 sin^5(x^2) + C

Is there anyway my answer can turn into the book's answer? I think I'm missing something major somewhere.
 
Have you tried \(\displaystyle \cos^{2}(x) = 1-\sin^{2}(x)\)?
 
Hello, hank!

Did you try tkhunny's suggestion?

I come up with: \(\displaystyle \:\frac{1}{10}\cos^4(x^2)\sin(x^2)\,+\,\frac{2}{15}\cos^2(x^2)\sin(x^2)\,+\,\frac{4}{15}\sin(x^2)\,+\,C\;\) Right!

According to the book, the answer is: \(\displaystyle \:\frac{1}{2}\sin(x^2)\,-\,\frac{1}{3}\sin^3(x^2)\,+\,\frac{1}{10}\sin^5(x^2)\,+\,C\)
Click to expand...

Let \(\displaystyle u \,=\,x^2\)

You have: \(\displaystyle \:\frac{1}{10}\cdot(1\,-\,\sin^2u)^2\cdot\sin u \,+\,\frac{2}{15}\cdot\left(1\,-\,\sin^2u\right)\cdot\sin u \,+\,\frac{4}{15}\cdot\sin u\)

Factor out \(\displaystyle \sin u:\;\;\sin u\cdot\bigg[\frac{1}{10}\cdot\left(1\,-\,2\cdot\sin^2u\,+\,\sin^4u\\right)\,+\,\frac{2}{15}\cdot\left(1\,-\,\sin^2u\right)\,+\,\frac{4}{15}\bigg]\)

. . \(\displaystyle =\;\sin u\cdot\bigg[\frac{1}{10}\,-\,\frac{1}{5}\sin^2u \,+\,\frac{1}{10}\sin^4u\,+\,\frac{2}{15}\,-\,\frac{2}{15}\sin^2u \,+\,\frac{4}{15}\bigg]\)

. . \(\displaystyle = \;\sin u\cdot\bigg[\left(\frac{1}{10}\,+\,\frac{2}{15}\,+\,\frac{4}{15}\right)\,+\,\left(-\frac{1}{5}\sin^2u\,-\,\frac{2}{15}\sin^2u\right) \,+\,\frac{1}{10}\sin^4u\bigg]\)

. . \(\displaystyle = \;\sin u\cdot\left[\frac{1}{2}\,-\,\frac{1}{3}\sin^2u\,+\,\frac{1}{10}\sin^4u\right]\)

. . \(\displaystyle =\;\frac{1}{2}\sin u \,-\,\frac{1}{3}\sin^3u \,+\,\frac{1}{10}\sin^5u\;\) . . . ta-DAA!

 
Ok, I see it now.

Duh, trig identity tripped me up.

Is there a way to do this without a reduction formula? Sort of like a nested trig sub?

Actually, I guess you could do a trig sub where you peel off one of the cosu and end up with something like:

1/2 * (cos^2(u))^2 * cos u du

From here you can do a trig identity for the cos^2(u) and go from there, right?

Actually, I just did it that way and it worked!

Thanks all..
 
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