# Math Help - Looking for an integration pointer!

1. ## 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.

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!

2. $\sin^{2}u = \frac {1}{2} - \frac {\cos 2u}{2}$

3. 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.
Put $t=\sqrt{x+1}$ written backwards as $t^2=x+1.$

4. 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,
$
\int_{0}^{\frac{\pi}{3}}\sin^2{(3x)}dx$

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

Therefore,
$
\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.

5. 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.

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

7. 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

EDIT- I'm mistaken, what you posted was 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 $\sin^{2} \Big(\frac{u}{2}\Big) = \frac {1-\cos u}{2}$

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

8. 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.

9. 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.

10. 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.

11. 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

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

$\frac {u}{6} \Big|^{\pi}_{0} - \frac{\sin(2u)}{12} \Big|^{\pi}_{0}$

12. 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!

13. 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 $\frac {1}{2} \int \cos v \ dv = \frac {1}{2} \sin v = \frac {1}{2} \sin(2u)$

14. 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.