## anonymous one year ago I have a test tmrw and I have NO CLUE how to solve these integrals! example: integrate dx/x(sqrt(x^2+16)) and integrate x^3(sqrt(9-x^2))dx can someone give me some help?

1. zepdrix

Oooo trig stuff :O fun!

2. Jhannybean

$\int \frac{dx}{x(\sqrt{x^2+16})}$ I believe this follows one of the basic integral forms of $$\sqrt{a^2+u^2}$$$\sf \int \frac{du}{u\sqrt{u^2+a^2}} = -\frac{1}{a}\ln \left|\frac{\sqrt{a^2+u^2}}{u}\right|+C$

3. anonymous

Am I supposed to be seeing an equation? sorry. on the second one I got

4. anonymous

integral u(sqrt(9-u))du where u=x^2 and 1/2du=xdx

5. zepdrix

Hmm I don't think that u-sub is going to be very effective. If you insist on doing a u-sub, you'll want to do $$\large\rm u=9-x^2$$ But trig sub would be a lot better :O

6. Jhannybean

hmm.. thats part of it. $\int x^3\sqrt{9-x^2}dx \qquad \implies \qquad \int\color{blue}{ x}\cdot \color{red}{x^2}\sqrt{9-\color{red}{x^2}}\color{blue}{dx}$$\color{red}{u=x^2}~,~ \color{blue}{du=2xdx \implies xdx =\dfrac{1}{2}du}$

7. anonymous

zepdrix (sorry Jhannybean) can you give me a push as to what a should be for the second problem?

8. zepdrix

Bahhh I take back what I said :) The u-sub I mentioned should work out really nicely! $$\large\rm u=9-x^2$$

9. Jhannybean

Ah ha... I just read up on the trig subs. $\sqrt{9-x^2} \qquad \implies\qquad \sqrt{a^2-x^2} ~,~ x=a\sin(\theta)$This!!! Now I recall.

10. zepdrix

You do that because you really would like to get rid of the subtraction under the root. $$\large\rm du=-2x~dx\qquad\to\qquad -\frac{1}{2}du=x~dx$$ But also:$\large\rm u=9-x^2\qquad\to\qquad x^2=u+9$ And then sub all your pieces in! :)$\large\rm \int\limits x^2\sqrt{9-x^2}~(x~dx)$

11. anonymous

Oh boy... Can anybody teach me how to set up the "triangle" to know to use cos, sin or tan?

12. Jhannybean

Therefore $\int x^3\sqrt{9-x^2}dx \qquad \implies \qquad \int(3\sin(\theta))^3\sqrt{9-(3\sin(\theta))^2 }\cdot 3\cos(\theta)d\theta$ Something like that?...

13. Jhannybean

Where $$x=3\sin(\theta) ~,\qquad ~ dx = 3\cos(\theta)d\theta$$

14. anonymous

ok... I only know how to do a trig subs with fractions where dx is on top...

15. Jhannybean

Wow,yeah I agree @zepdrix , this is going to be really long x_x haha crap.

16. zepdrix

I'll group the steps together that I posted earlier, maybe it will help :o

17. zepdrix

$\large\rm \int\limits\limits x^3\sqrt{9-x^2}~dx=\int\limits \color{#F35633}{x^2}\sqrt{\color{#CC0033}{9-x^2}}~(\color{green}{x~dx})$Making the substitution:$\large\rm \color{#CC0033 }{u=9-x^2}$Adding x^2 to each side, subtracting u from each side gives us,$\large\rm \color{#F35633}{x^2=9-u}$Differentiating the red one gives us,$\large\rm du=-2x~dx\qquad\to\qquad \color{green}{-\frac{1}{2}du=dx}$$\large\rm \int\limits\limits \color{#F35633}{x^2}\sqrt{\color{#CC0033}{9-x^2}}~(\color{green}{x~dx})=\int\limits\limits \color{#F35633}{(9-u)}\sqrt{\color{#CC0033}{u}}~\left(\color{green}{-\frac{1}{2}du}\right)$

18. zepdrix

From there it's not too bad, ya? :) Rewrite your square root as a 1/2 power,$\large\rm =-\frac{1}{2}(9-u)u^{1/2}~du$Distribute the u^(1/2), and then just apply your power rule to each term.

19. zepdrix

Doh :d typo on the green step, my bad$\large\rm du=-2x~dx\qquad\to\qquad \color{green}{-\frac{1}{2}du=x~dx}$

20. anonymous

HEY! What if I set u^2=9-x^2? because then the sqrt is resolved to just u, and I'm left with integrate: (9-u^2)(-u^2)du?

21. zepdrix

That's a really strange way to approach it :D But I think it works! Let's see...$\large\rm u^2=9-x^2,\qquad\quad x^2=u^2+9,\qquad\quad u=\sqrt{9-x^2}$Differentiating the first part,$\large\rm 2u~du=-2x~dx\qquad\to\qquad -u~du=x~dx$Subbing everything in gives you,$\large\rm \int\limits x^2\sqrt{9-x^2}~(x~dx)=\int\limits (u^2+9)\cdot u\cdot (-u~du)$I think maybe your sign is off in the middle, looks good besides that though.

22. zepdrix

23. anonymous

I worked it out both waym and they both check with the answer in the book! Any help on the first one? I know it has to do with x=sina or something?

24. zepdrix

$\large\rm \int\limits \frac{dx}{x \sqrt{x^2+16}}$You would really like to get $$\large\rm (stuff)^2+1$$ under the root, then you could make the substitution:$\large\rm stuff=\tan\theta$And we would get,$\large\rm (\tan \theta)^2+1=\sec^2\theta$Trig sub is a clever way to get rid of the addition/subtraction between terms! It allows us to easily take this out of the root.

25. zepdrix

How do we get +1 though? Hmm we have a +16... Well, if we can make a 16 show up in our "stuff", then we can factor a 16 out of each term, leaving us with 16(stuff^2+1) under the root.

26. zepdrix

Hopefully that makes some sense :p I know I'm throwing the word stuff around a lot...

27. zepdrix

So it seems that sine isn't the sub we're looking for. Since we have addition, we're going to head over to tangent, since the square identity involving tangent has addition in it.

28. anonymous

Looking around, I found this helpful powerpiont. It gives a good start to anyone looking at this thread when its archived: https://drive.google.com/file/d/0B0lXunoGraQaQ2tfZ0NCbFpUeDQ/edit

29. anonymous

Speaking of, we need an a(tan)theta = x for this one.

30. zepdrix

lol

31. anonymous

give me a second to work this out a bit...

32. Astrophysics

Did you work it out? I like this question so I'll show a solution you can then compare your answer to.. $\int\limits \frac{ dx }{ x \sqrt{x^2+16} }$ $x= \tan \theta \implies dx = \sec^2\theta d \theta$ that will be our substitution, so our new integral will look like $\int\limits \frac{ 4\sec^2\theta d \theta }{ \tan \theta \sqrt{(4\tan \theta)^2+16} }$ now notice in the denominator we can factor out the 16, $\int\limits \frac{ 4 \sec^2 \theta d \theta }{ 4\tan \theta \sqrt{16 \tan^2 \theta+16} } \implies \int\limits \frac{ \sec^2 \theta d \theta }{ 4 \tan \theta \sqrt{16(\tan^2 \theta+1)} }$ now we can use the following identity $\tan^2 \theta +1 = \sec^2 \theta$ so we get $\int\limits \frac{ \sec^2 \theta d \theta }{ \tan \theta \sqrt{16 \sec^2 \theta} } \implies \frac{ 1 }{ 4 }\int\limits \frac{ \sec^2 \theta }{ \tan \theta \sec \theta }d \theta$$\frac{ 1 }{ 4 } \int\limits \frac{ \sec \theta d \theta }{ \tan \theta } d \theta \implies \frac{ 1 }{ 4 } \int\limits \csc \theta d \theta$ and remember the integral of csc theta is $\frac{ 1 }{ 4 } \int\limits \csc \theta d \theta = \frac{ 1 }{ 4 } \ln|\csc \theta-\cot \theta|+C$ but we're not exactly done yet as we made a substitution earlier. |dw:1443826146119:dw| looking at the trig ratios we finally get $\frac{ 1 }{ 4 }(\ln|x|-\ln|\sqrt{x^2+16}+4|)+C$