Simple indefinite integration

**3n1gm4** · May 26th 2008, 10:54 AM

Hi all, this is my first post

I'd like to know how to solve this integral in dx (i dont know how to use the forum tag MATH and if there is the integration symbol... sorry) step by step.

$x (2x-x^2)^(1/2)$
it is sqrt(2x-x^2), but i dont know the syntax of the tag so I cant writeis as it should be... sorry again :\

I tried but I failed

Some website solve it but they dont give me the step by step solution... anyone can (at least) tell me what method I should use?

Thank you

EDIT:
Using the correct syntax is:

$\displaystyle\int x\sqrt{2x-x^2} \, \mathrm{d}x<br />$

I'm starting to learn the syntax

**galactus** · May 26th 2008, 12:11 PM

Here's one way. Someone will probably have a more efficient method, but I am sure there's worse.

$\int{x\sqrt{2x-x^{2}}}dx$

Let $x=2cos^{2}(t), \;\ dx=-4sin(t)cos(t)dt$

Make the subs and we get:

$-16\int{sin^{2}(t)cos^{4}(t)}dt$

Now, for this. We'll attach the -16 later:

$\int{sin^{2}(t)cos^{4}(t)}dt$

$=\frac{1}{8}\int{(1-cos(2t))(1+cos(2t))^{2}}dt$

$=\frac{1}{8}\int{(1-cos^{2}(2t))(1+cos(2t))}dt$

$=\frac{1}{8}\int{sin^{2}(2t)}dt+\frac{1}{8}\int{si n^{2}(2t)cos(2t)}dt$

$=\frac{1}{16}\int{(1-cos(4t))}dt+\frac{1}{48}sin^{3}(2t)$

$=\frac{1}{16}t-\frac{1}{64}sin(4t)+\frac{1}{48}sin^{3}(2t)$

Now, multiply by the -16:

$-t+\frac{1}{4}sin(4t)-\frac{1}{3}sin^{3}(2t)+C$

Now, sub in $t=cos^{-1}(\sqrt{\frac{x}{2}})$ and it is back in terms of x

**3n1gm4** · May 26th 2008, 01:28 PM

Thank you so much!

Can you tell me how did you "see" that $2cos^{2}(t)$ was the right sub? I mean, you did not try, what I missed?

Ah... and can you explain me the step from
$\int{sin^{2}(t)cos^{4}(t)}dt$
to
$<br /> =\frac{1}{8}\int{(1-cos(2t))(1+cos(2t))^{2}}dt<br />$

Thank you again

**galactus** · May 26th 2008, 01:57 PM

How did I 'see' it?. I don't know, I just did. I seen when I subbed that into the function to integrate I got: $4sin(t)cos^{3}(t)$ , then I tacked on the derivative portion and got what we integrated.

As for the identities. They came from:

$sin^{2}(t)=\frac{1-cos(2t)}{2}; \;\ cos^{2}(t)=\frac{1+cos(2t)}{2}$

See a little clearer now?.

**3n1gm4** · May 26th 2008, 02:03 PM

Originally Posted by galactus

See a little clearer now?

Yes a bit. It works, but the solution websites gave me is different :O
I cant check that solution because there are no steps, I don't see any error in your integration... i'll try derive both of them...

**galactus** · May 26th 2008, 02:19 PM

If you set it back in terms of x by making the substitution $t=cos^{-1}(\sqrt{\frac{x}{2}})$ , you should get something equivalent, though it may be difficult to see.

**3n1gm4** · May 26th 2008, 03:12 PM

Originally Posted by galactus

If you set it back in terms of x by making the substitution $t=cos^{-1}(\sqrt{\frac{x}{2}})$ , you should get something equivalent, though it may be difficult to see.

I see, but is the logarithm that's scaring me

I cant figure where it comes from :\

**galactus** · May 26th 2008, 05:00 PM

Well, I get

$\frac{\sqrt{x}(2x^{2}-x-3)\sqrt{2-x}-6cos^{-1}(\frac{\sqrt{2x}}{2})}{6}$

Which, believe it or not, gives the same result.

Sometimes technology gives cumbersome, yet equivalent, solutions to a problem.

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