Looking for an integration pointer!

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July 11th 2009, 12:29 PM
superduper

Looking for an integration pointer!

Hey guys, I got two problems here, that are probably extremely basic; but i'm having a hard time with these. When I finally find a $u$ -substitution path I can generally work my way through, but I'm hitting a block on these, even just a pointer or what $u$ should be would be great!

1) $\int x^2 \sqrt{x+1}dx$

For this one, I'm trying to find a $u$ sub but I can't seem to make it work, I keep wishing real hard that the $x^2$ was under the radical and the $x$ was outside.(Speechless)

2) $\int\sin^2{(3x)}dx$
$= \int(\sin{(3x)})^2dx$

Again, no idea what to set $u$ equal to because there is no $2(\sin{(3x)})(\cos{(3x)})(3)$ with the $dx$

Any toss as to where to go next would be great, thanks for taking a look!
July 11th 2009, 12:38 PM
Random Variable

$\sin^{2}u = \frac {1}{2} - \frac {\cos 2u}{2}$
July 11th 2009, 12:41 PM
Krizalid

Quote:

Originally Posted by superduper

1) $\int x^2 \sqrt{x+1}dx$

For this one, I'm trying to find a $u$ sub but I can't seem to make it work, I keep wishing real hard that the $x^2$ was under the radical and the $x$ was outside.(Speechless)

Put $t=\sqrt{x+1}$ written backwards as $t^2=x+1.$
July 12th 2009, 02:32 PM
superduper

Hey, so I got an answer for the radical integrand, but I'm still having trouble with the Trig question.

I set $u=3x$ , and to integrate my textbook has a formula that states $\int\sin^2{u}du=\frac{1}{2}u-\frac{1}{4}\sin{2u}+C$ .

So I applied this as follows,
$<br /> \int_{0}^{\frac{\pi}{3}}\sin^2{(3x)}dx$
Where $u=3x$ and $du=3dx$

Therefore,
$<br /> \int_{0}^{\frac{\pi}{3}}\sin^2{(3x)}dx$
= $3\int_{0}^{\frac{\pi}{3}}\sin^2{(u)}du$
= $3(\frac{1}{2}u+\frac{1}{4}\sin{(2u)})|^{\pi/3}_{0}$
= $[\frac{3}{2}\sin{((3)(\frac{\pi}{3}))}+\frac{3}{4}\ sin{((2)(3)(\frac{\pi}{3})}]-[\frac{3}{2}\sin{(0)}+\frac{3}{4}\sin{(0)}]$
= $[\frac{3}{2}\sin{(\pi)}+\frac{3}{4}\sin{(2\pi)}]-[\frac{3}{2}\sin{(0)}+\frac{3}{4}\sin{(0)}]$
= $[0+0+-[0+0]$ ?

I'm not sure where I'm going wrong here.

I was going to sub in the post that RandomVariable made, but I was just wondering if you could clarify that for me, I thought that identity was supposed to look like

$sin^2{\frac{u}{2}}=\frac{1+cos(u)}{2}$

EDIT- I'm mistaken, what you posted was $cos2u$ but I'm still lost as to where that identity comes from? I can't find it anywhere in my textbook.

Thanks again for the help guys.
July 12th 2009, 02:44 PM
Plato

Go to this website W|A
Type in this exact phrase: integrate x^2 sqrt[x+1].
Click the equals bar at the end if the input window.
Be sure you click ‘show steps’ to see the solution.
July 12th 2009, 03:02 PM
superduper

Wow Plato, that website is amazing, thanks for showing me that!

That solution showed me that I was wrong, but I was on the right track, messed up the u's somewhere.

Right off the bat, WolframAlpha says

$\int x^2\sqrt{x+1}dx$

Where $u=\sqrt{x+1}$ and $du=\frac{1}{2\sqrt{x+1}}$

So they say

$\int x^2\sqrt{x+1}dx$
= $2\int u^2 (u^2-1)^2 du$
= $2\int (u^6-2u^4+u^2)du$

Could you elaborate on this for me please? From the above relation shouldn't $u^2=x+1$ not $u^2=x^2$ I have no idea how they got this, I came up with

$\int_{0}^3 (u-1)^2\sqrt{u}du$ where $u=x+1$
= $\int_{0}^3 (x^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}})du$

Thanks for your time!
July 12th 2009, 03:25 PM
Random Variable

Quote:

I was going to sub in the post that RandomVariable made, but I was just wondering if you could clarify that for me, I thought that identity was supposed to look like

http://www.mathhelpforum.com/math-he...58cacba3-1.gif

EDIT- I'm mistaken, what you posted was http://www.mathhelpforum.com/math-he...e194f001-1.gif but I'm still lost as to where that identity comes from? I can't find it anywhere in my textbook.

Just let u= 2x. And it should be $\sin^{2} \Big(\frac{u}{2}\Big) = \frac {1-\cos u}{2}$

$\cos^{2} \Big(\frac {u}{2} \Big) = \frac {1+\cos u}{2}$
July 12th 2009, 03:32 PM
superduper

But the original equation was $sin(3x)$ so if we make $u=2x$ don't we have to turn that into $\frac{3}{2}u=3x$ or something to make that equation usable? Or actually $\frac{2}{3}u=\frac{2}{3}(3x)$ ? I'm getting lost now.
July 12th 2009, 04:25 PM
Random Variable

Quote:

Originally Posted by superduper

But the original equation was $sin(3x)$ so if we make $u=2x$ don't we have to turn that into $\frac{3}{2}u=3x$ or something to make that equation usable? Or actually $\frac{2}{3}u=\frac{2}{3}(3x)$ ? I'm getting lost now.

$\sin^{2}(3x) = \frac {1 - \cos2(3x)}{2} = \frac {1-\cos(6x)}{2}$

The u=2x thing was to show that the identity you remember is the same identity.
July 12th 2009, 04:51 PM
superduper

Oooh, that makes sense. Thanks for the clarification!

So here's what I have so far, would you mind poking over it and letting me know what you think?

$\int_{0}^{\frac{\pi}{3}} sin^2(3x)dx$

Where $u=3x$ and $3du=dx$

Therefore,

$\int_{0}^{\frac{\pi}{3}} sin^2(3x)dx$
= $3\int_{0}^{\frac{\pi}{3}} sin^2u du$
= $3\int_{0}^{\frac{\pi}{3}} \frac{1-cos(2u)}{2} du$
= $\frac{3}{2} \int_{0}^{\frac{\pi}{3}} 1-cos(2u) du$
= $\frac{3}{2}(u-sin(2u))|_{0}^{\frac{\pi}{3}}$

Looking good so far? Thanks again for your time.
July 12th 2009, 05:16 PM
Random Variable

if you let $u = 3x$ , then $du=3dx$ which implies that $\frac {du}{3} = dx$

so the integral becomes $\frac {1}{3} \int \sin^2(u) \ du$

Now you need to change the limits of integration.

The upper limit, which was $x=\frac{\pi}{3}$ , becomes $u =3*\frac{\pi}{3} = \pi$

The lower limt, which was $x=0$ , becomes $u =3*0=0$

so the integral now becomes $\frac{1}{3}\int^{\pi}_{0} \sin^{2}(u) \ du$

now using the identity

$= \frac{1}{3} \int^{\pi}_{0} \big(\frac {1}{2} - \frac{\cos(2u)}{2} \Big) du$ $= \frac{1}{6}\int^{\pi}_{0} du - \frac{1}{6} \int^{\pi}_{0} \cos(2u) \ du$

$\frac {u}{6} \Big|^{\pi}_{0} - \frac{\sin(2u)}{12} \Big|^{\pi}_{0}$
July 12th 2009, 05:28 PM
superduper

Perfect, I can finish with that. Just one quick question on the last step.

Why is $\int cos(2u)du=\frac{sin(2u)}{2}$ ? I don't see that formula anywhere either, I thought it was;

$\int cos (u) du= sin (u)$

Thanks!
July 12th 2009, 05:36 PM
Random Variable

Quote:

Originally Posted by superduper

Perfect, I can finish with that. Just one quick question on the last step.

Why is $\int cos(2u)du=\frac{sin(2u)}{2}$ ? I don't see that formula anywhere either, I thought it was;

$\int cos (u) du= sin (u)$

Thanks!

You have to make a substituion.

Let $v = 2u$ , then $dv=2du$ which implies $\frac {dv}{2}=du$

so the intergal becomes $\frac {1}{2} \int \cos v \ dv = \frac {1}{2} \sin v = \frac {1}{2} \sin(2u)$
July 12th 2009, 05:37 PM
superduper

Oh I see, that makes sense, I've never done more than 1 substitution in the same problem before. Thanks for sticking with me here, really appreciate it.