I know I am messing up something with the substitutions but I am not sure what.

$$\int \sqrt{1-4x^2}$$

$$u = 4x, du = 4 \,dx$$

$$\frac{1}{4}\int \sqrt{1-u^2}$$

$u = \sin \theta$

$$\frac{1}{4}\int \sqrt{1-\sin^2 \theta} = \frac{1}{4}\int \sqrt{ \cos^2 \theta} = \frac{1}{4}\int \cos \theta = \frac{\sin \theta}{4}$$

Replace $\theta$ with $u$

$$\frac{\sin (\arcsin u)}{4} = \frac{u}{4} = \frac{4x}{4} = x$$

This is wrong and I have no idea why.

  • 3
    $\begingroup$ It is $u=2x$. So we get $du=2\,dx$ and end up with $\int 2\sqrt{1-u^2}\,du$. Then $u=\sin\theta$. Remember (you have been doing so lately, but slipped) that $du=\cos\theta\,d\theta$. Please don't forget to write integrals properly. The substitution went wrong probably because you wrote $\int\sqrt{1-u^2}$ instead of $\int\sqrt{1-u^2}\,du$. I know, that $d$ stuff is a nuisance, but it helps protect against error. $\endgroup$ – André Nicolas Jun 4 '12 at 17:26
  • $\begingroup$ By your substitution $u=4x$, you have $$\begin{equation*} \int \sqrt{1-4x^{2}}dx=\int \sqrt{1-4\frac{1}{4^{2}}u^{2}}\,\frac{1}{4} du=\int \frac{1}{4}\sqrt{1-\frac{1}{4}u^{2}}\,du \end{equation*}$$ $\endgroup$ – Américo Tavares Jun 4 '12 at 17:26
  • $\begingroup$ Don't forget the $+C$! $\endgroup$ – TMM Jun 4 '12 at 17:35
  • $\begingroup$ I do not follow this at all. $\endgroup$ – toby yeats Jun 4 '12 at 17:37
  • 1
    $\begingroup$ I do not understand how to manipulate equations through 2 sets of variables. $\endgroup$ – toby yeats Jun 4 '12 at 18:00

$$\int \sqrt{1-4x^2}\ dx$$ Substitute $$\begin{align*}x &= \frac{\sin (\theta)}{2}\\ dx &= \frac{\cos(\theta)}{2}d \theta\end{align*}$$ So the intregral become $$\begin{align*} \int \sqrt{1-4x^2}\ dx &=\frac{1}{2}\int \cos^2(\theta)d\theta\\ &=\frac{1}{2}\int \frac{\cos(2\theta) + 1}{2}d\theta\\ &=\frac{\sin(2\theta)}{8}+\frac{\theta}{4} \end{align*}$$ Again $$\theta = \arcsin(2x)$$ And $$\begin{align*} \sin 2\theta &= 2\sin \theta \cos \theta\\ &=4x\sqrt{1-4x^2} \end{align*}$$ So the answer is $$\frac{x\sqrt{1-4x^2}}{2}+\frac{\arcsin(2x)}{4}+C$$

  • $\begingroup$ I do not see how this works, it looks like have sin/2 will not work. How does that work? Also I do not follow the part after "So the integral become" it looks like you forgot to take the square root. $\endgroup$ – toby yeats Jun 4 '12 at 18:15
  • $\begingroup$ @Jordan see i have used sin(x)/2 so that inside squareroot 4 will vanish and we will be left with cos(x) and another cos(x)/2 will come when we $dx = \cos(\theta)/2 d \theta$ hence the $cos^2(\theta)/2$ $\endgroup$ – Saurabh Jun 4 '12 at 18:21
  • $\begingroup$ @Jordan with that change of variables: $$\sqrt{1-4x}dx=\sqrt{1-4\left(\frac{\sin\theta}{2}\right)^2}\frac{\cos\theta}{2}\ d\theta,$$ can follow from here? $\endgroup$ – leo Jun 4 '12 at 18:22
  • $\begingroup$ @SaurabhHota nice answer. Please don't forget to add a constant to the result. $\endgroup$ – leo Jun 4 '12 at 18:23
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    $\begingroup$ @Jordan It is as is because: $$\begin{align*} \sqrt{1-4x}dx&=\sqrt{1-4\left(\frac{\sin\theta}{2}\right)^2}\frac{\cos\theta}{2‌​}\ d\theta\\ &= \sqrt{1-4\cdot\frac{\sin^2\theta}{4}}\frac{\cos\theta}{2}\ d\theta\\ &=\sqrt{1-\sin^2\theta}\frac{\cos\theta}{2}\ d\theta \end{align*}$$ and you know what $\sqrt{1-\sin^2\theta}$ is, right? $\endgroup$ – leo Jun 4 '12 at 18:31

You should substitute $u=2x$, to get the next integral:

$$\int du/2 \sqrt{1-u^2}$$

Now change variables to $u=\cos(t)$ to get:

$$-\int (\sin(t)/2) \sin(t) dt$$ which you solve by parts or by using the formula for $\cos(2\theta)$.

Hope I helped.

  • $\begingroup$ I do not quit follow what happened here. Why would I use 2x instead of 4x? That seems counter intuitive. $\endgroup$ – toby yeats Jun 4 '12 at 17:21
  • $\begingroup$ You want the $4x^2$ to be replaced by $u^2$, so you want $4x^2=u^2$. To solve for $u$, take square roots of both sides to get $u=2x$. $\endgroup$ – John Engbers Jun 4 '12 at 17:26
  • $\begingroup$ I have no idea what is happening, where did the du go? $\endgroup$ – toby yeats Jun 4 '12 at 17:38
  • $\begingroup$ @Jordan better question: where did the du go in your attempt? The du here was transformed into $-\sin(t)dt$. $\endgroup$ – Robert Mastragostino Jun 4 '12 at 18:11

Some things to watch out for:

  1. As already mentioned, if you substitute $u = 4x$, then you get $u^2 = (4x)^2 = 16x^2$ instead of $u^2 = 4x^2$. If you use $u = 2x$, then $u^2 = (2x)^2 = 4x^2$.
  2. An annoying but important part of practicing good mathematics is being precise.

    • In your integrals, you forgot the $dx$ and $du$ several times, which also lead you to forget to express $du$ in terms of $d\theta$. Perhaps you have heard of Riemann integrals; these may help you understand why the "$dx$" is there.
    • You forgot the fact that if $F'(x) = f(x)$, then $\int f(x) dx = F(x) + C$. After all, when taking derivatives, all constants disappear, so the derivatives of $F(x)$ and $F(x) + 5$ are exactly the same. Thus, when calculating indefinite integrals, you do not know the constant term of the resulting function. So we generally write $\int f(x) dx = F(x) + C$ for some unknown constant $C$.

Here's a step-by-step approach of how I would do it. First, substitute $u = 2x$, so that $u^2 = (2x)^2 = 4x^2$, and apply $du = 2dx$:

$$\int \sqrt{1 - \color{red}{4x^2}} \color{blue}{dx} = \int \sqrt{1 - \color{red}{u^2}} \color{blue}{\frac{du}{2}} = \frac{1}{2} \int \sqrt{1 - u^2} du.$$

Then substitute $u = \sin \theta$ and $du = \cos \theta d\theta$ to get:

$$\frac{1}{2} \int \sqrt{1 - \color{red}{u^2}} \color{blue}{du} = \frac{1}{2} \int \sqrt{1 - \color{red}{\sin^2 \theta}} \color{blue}{\cos \theta d\theta} = \frac{1}{2} \int \sqrt{\cos^2 \theta} \cos \theta d\theta = \frac{1}{2} \int \cos^2 \theta d\theta.$$

Next, apply the double angle formula for the cosine, $\cos^2 \theta = \frac{1}{2} (1 + \cos 2\theta)$, to get

$$\frac{1}{2} \int \color{red}{\cos^2 \theta} d\theta = \frac{1}{2} \int \color{red}{\frac{1 + \cos 2\theta}{2}} d\theta = \frac{1}{4} \int (1 + \cos 2\theta) d\theta.$$

Now split in two integrals and calculate those separately:

$$\frac{1}{4} \int (\color{red}{1} + \color{blue}{\cos 2\theta}) d\theta = \frac{1}{4} \left(\int \color{red}{1} d\theta + \int \color{blue}{\cos 2\theta} d\theta\right) = \frac{1}{4}\left(\color{red}{\theta + C_1} + \color{blue}{\frac{\sin 2\theta}{2} + C_2}\right) \\ = \frac{\theta}{4} + \frac{\sin 2\theta}{8} + C.$$

Applying a double angle formula for the sine, $\sin 2\theta = 2 \sin \theta \cos \theta$, we get

$$\frac{\theta}{4} + \frac{\color{red}{\sin 2\theta}}{8} + C = \frac{\theta}{4} + \frac{\color{red}{2 \sin \theta \cos \theta}}{8} + C = \frac{\theta}{4} + \frac{\sin \theta \cos \theta}{4} + C.$$

Substituting back $\theta = \arcsin u$, and using $\cos (\arcsin u) = \sqrt{1 - u^2}$, we get

$$\frac{\color{red}{\theta}}{4} + \frac{\sin \color{blue}{\theta} \cos \color{green}{\theta}}{4} + C = \frac{\color{red}{\arcsin u}}{4} + \frac{\sin(\color{blue}{\arcsin u}) \cos(\color{green}{\arcsin u})}{4} + C = \frac{\arcsin u}{4} + \frac{u \sqrt{1 - u^2}}{4} + C.$$

Finally substituting $u = 2x$, you get

$$\frac{\arcsin \color{red}{u}}{4} + \frac{\color{blue}{u} \sqrt{1 - \color{green}{u}^2}}{4} + C = \frac{\arcsin \color{red}{2x}}{4} + \frac{\color{blue}{2x} \sqrt{1 - \color{green}{(2x)}^2}}{4} + C = \boxed{\displaystyle\frac{\arcsin 2x}{4} + \frac{x \sqrt{1 - 4x^2}}{2} + C}.$$

  • $\begingroup$ Why is cosarcsin u equal to that statement? $\endgroup$ – toby yeats Jun 4 '12 at 19:26
  • $\begingroup$ @Jordan See for instance the paragraph "But... what is $\cos(\arcsin(u))$?" in this answer to another question. $\endgroup$ – TMM Jun 4 '12 at 19:37

Let $2x = u$. So, $dx = \frac{1}{2}du$ $$\frac{1}{2}\int \sqrt{1-u^2} du$$ Replace $u = \sin \theta$, $du = cos \theta\ d\theta$

So, equation will be
$$\frac{1}{2}\int \sqrt{1-sin^2 \theta}\ cos\theta\ d\theta$$ $$\frac{1}{2}\int \sqrt{\cos^2\theta}\ cos\theta\ d\theta$$ $$\frac{1}{2}\int cos^2\theta\ d\theta$$ Add and subtract $\frac{1}{4}\int d\theta$ $$\frac{1}{4}\int 2cos^2\theta\ d\theta - \frac{1}{4}\int d\theta + \frac{1}{4}\int d\theta$$ $$\frac{1}{4}\int (2cos^2\theta-1)\ d\theta+ \frac{1}{4}\int d\theta$$ $$\frac{1}{4}\int (2cos^2\theta-1)\ d\theta+ \frac{1}{4}\int d\theta$$ $$\frac{1}{4}\int \cos2\theta\ d\theta+ \frac{1}{4}\int d\theta$$ $$\frac{1}{8} (\sin2\theta) + \frac{1}{4} \theta + C$$ $$\frac{1}{8} (2\sin\theta.\cos\theta) + \frac{1}{4} \theta + C$$ As u = $\sin\theta$ $$\frac{1}{4} (u\sqrt{1-u^2}) + \frac{1}{4} \sin^{-1}u + C$$ As $2x = u$ $$\frac{1}{4} (2x\sqrt{1-4x^2}) + \frac{1}{4} \sin^{-1}2x + C$$ $$\frac{1}{2} (x\sqrt{1-4x^2}) + \frac{1}{4} \sin^{-1}2x + C$$

  • $\begingroup$ What happened to the 4? $\endgroup$ – toby yeats Jun 4 '12 at 17:56
  • $\begingroup$ @Jordan : Just wait for a second, I am not completed yet $\endgroup$ – rekenerd Jun 4 '12 at 18:00
  • $\begingroup$ @Jordan : 4 in which step? $\endgroup$ – rekenerd Jun 4 '12 at 18:04

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