2

I am trying to find $$\int \frac {dx}{x^2 \sqrt{4-x^2}}$$

I make $t=2\sin x$

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

$$\int \frac {dx}{x^2 \sqrt{4(1-\sin^2 t)}}$$

$$\int \frac {dx}{x^2 \sqrt{4(\cos^2 t)}}$$

$$\int \frac {dx}{x^2 \cdot 2\cos t}$$

I do not really know where to go from here. I have two variables and that is really bad but I do not know how to write $x$ in terms of $t$.

Américo Tavares
  • 38,602
  • I'm afraid to say @Jordan that it already looks as if you only want someone here to solve your homework, without you trying/succeeding to get some benefit of this all (and I believe you when you say you don't like this but you only have to do it). I think that either you try *really harder* or else try to get some personalised help at your college/H.S. from your instructor. I, for one, can't continue to answer your questions until I see some real progress in you. – DonAntonio Jun 03 '12 at 22:43
  • 1
    @DonAntonio: He's certainly making progress. He added his solution to the questions today to find out what he has done wrong. If you can't continue to answer his questions, just don't answer his questions. – Gigili Jun 03 '12 at 22:51
  • Yes @Gigili, that's what I said: I can't. Wasn't this clear enough to you? – DonAntonio Jun 03 '12 at 23:50
  • It was clear enough that I replied "don't" @DonAntonio. – Gigili Jun 03 '12 at 23:52

2 Answers2

3

You must replace every occurrence of $x$. Let $x=2\sin t$. This is so that the square root of $4-x^2$ will be nice. Then $dx=2\cos t\,dt$, and $\sqrt{4-x^2}=2\cos t$. So we end up needing to find $$\int \frac{2\cos t}{(4\sin^2 t)(2\cos t)}dt.$$ There is cancellation, and we end up needing to find $\int\frac{1}{4}\csc^2 t\,dt$.

André Nicolas
  • 507,029
  • I do not see what you did to solve for x. –  Jun 03 '12 at 22:49
  • We said let $2 \sin t$. So there is no need to solve for $x$. – André Nicolas Jun 03 '12 at 22:54
  • 1
    Oh I get it, I have no idea how I couldn't figure that out on my own. Also I have no idea how to get csc, I think I need to turn it back into sin but I am not sure. –  Jun 03 '12 at 22:55
  • I have no idea what to do next, my book gets a really weird answer. Do I plug back in t or do I find the anti derivative? –  Jun 03 '12 at 23:14
  • I am going through my book trying to understand the last part with the triangle and I just do not understand it at all. –  Jun 03 '12 at 23:30
  • @Jordan, what is the derivative of $\cot x$? Can you use that information in finding this integral? I'd recommend writing out the derivatives of all trig functions $\sin,\cos,\tan,\sec,\csc,\cot$ and having them with you on paper as you do the problems, so that you can more easily recognize where you need them. – Robert Mastragostino Jun 04 '12 at 00:18
  • @RobertMastragostino I know that the derivative of cot is -csc^2 but that doesn't help me get an answer. –  Jun 04 '12 at 00:23
  • 1
    @Jordan, integration is the reverse of differentiation, and $cot$ is so close! Can you think of what to multiply $cot$ by, so that when differentiated you get $1/4\csc^2$? Because that (plus the constant) will be your answer. – Robert Mastragostino Jun 04 '12 at 00:29
  • @Robert I do not understand why that will be my answer. –  Jun 04 '12 at 00:35
  • instead of differentiating $cot$, differentiate $-1/4cot$ instead. The point is that multiplying by a constant isn't affected by derivatives and integrals, so if you ever see a close solution that's off by a constant you can just nudge it slightly to get the answer you need. I really suggest doing less complicated examples. You need to have the basic integral properties understood like the back of your hand before you attempt more complicated questions like these. – Robert Mastragostino Jun 04 '12 at 00:52
3

On many occasions if you see a factor of $x^2$ in the denominator of the integrand, you may find it convenient to get rid of it by substituting: $$x=\frac{1}{t}$$ If you are computing a definite integral, take care if the origin is on the integration interval. Hence $$dx=-\frac{dt}{t^2}$$ $$I=-\int\frac{t^{2}t \, dt}{t^{2}\sqrt{4t^{2}-1}}=-\frac{1}{8}\int\frac{d\left(4t^{2}-1\right)}{\sqrt{4t^{2}-1}}=-\frac{1}{8}2\sqrt{4t^{2}-1}=-\frac{\sqrt{4-x^{2}}}{x}$$ I have used the fact that $dx=\frac{1}{a} \, d(ax+b)$, ($a\ne0$) and $(\sqrt{t})'=\frac{1}{2\sqrt{t}}$

Community
  • 1
Valentin
  • 4,563
  • 19
  • 22