Can someone help me find the integrals for the ff:

$$\int \frac{\sqrt{9z^{2}-1}}{27z^{4}}\,dz$$

$$\int y^{2}\sqrt{25-y^{2}} dy$$

For the second one I have tried to do answer it and this is what I have made:

$x=5\sin(\theta) $


$\int 25\sin^{2}\theta \sqrt{25-25\sin^{2}\theta }\cdot 5\cos\theta\, d\theta$

$625\int \sin^{2}\theta \cos^{2}\theta\, d\theta $

$625\int (1-\cos^{2}\theta ) \sin^{2}\theta \cos^{2}\theta\, d\theta $

and at this point i'm not sure if I did it correct or not.


$$\int \frac{dx}{x^{2}{\sqrt{4x^{2}-9}}}$$

  • $\begingroup$ Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: we want to see that you have put significant work into the problem. Also, you left out the $dz$ in your first integral. I just put it in. $\endgroup$ – Rory Daulton Feb 28 '16 at 13:28
  • $\begingroup$ Yep thank you, I must have missed it sorry. Oh, I have finished answering the second one already. The third one seems a little bit off for me since I'm not sure where to start. Because I don't know how to exactly or what to use if you have a problem like that. Sorry for that. $\endgroup$ – user318463 Feb 28 '16 at 13:49
  • $\begingroup$ I think i made a wrong thing about the 2. this is what i did. x=5sin(Theta) dx=5cos(theta) d(theta) link and ended up having this one link $\endgroup$ – user318463 Feb 28 '16 at 14:02


  1. Take $z=\frac13\sec\theta$

  2. Take $y=5\sin\theta$

  3. Take $x=\frac32\sec\theta$

  • $\begingroup$ Oh! I'm doing number 2 right now and it seems like i'm in the right direction. But for the first one I cannot put out the one you gave. How does it happened? $\endgroup$ – user318463 Feb 28 '16 at 13:35
  • $\begingroup$ I mean I don't know where did the 1/3 came from. $\endgroup$ – user318463 Feb 28 '16 at 13:47
  • $\begingroup$ I would love to know how you think this substitutions, please $\endgroup$ – DonAntonio Feb 28 '16 at 13:50
  • $\begingroup$ @user318463 The idea is to simplify the numerator by removing the square root symbol. It is of the form $ax^2-1$. It is also known that $\sec^2\theta-1=\tan^2\theta$. This gives the idea of trying this substitution. Just using $\sec\theta$ won't work, as you get $9\sec^2\theta-1$, which doesn't simplify. You need to look for a constant to multiply with it to cancel out the $9$, and that is $\frac13$. $\endgroup$ – GoodDeeds Feb 28 '16 at 13:50
  • $\begingroup$ @Joanpemo The motivation behind the substitution is to remove the square root sign to simplify our expression. In trigonometry, it is known that $\sin^2\theta+\cos^2\theta=1$,$\sec^2\theta-\tan^2\theta=1$ and $\csc^2\theta-\cot^2\theta=1$. So, whenever an expression of the form $ax^2-b$ or $ax^2+b$(where $a,b$ are constants) occurs under the square root sign, it gives an idea to try out the appropriate trigonometric substitution. $\endgroup$ – GoodDeeds Feb 28 '16 at 13:54

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