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I have to calculate $\int \frac{1}{\sqrt{1 - x^2}} \operatorname{d}x$ forwards, using known rules like partial integration or substitution.

What I'm not allowed to do is simply show that $\frac{\operatorname{d}}{\operatorname{d} x} \arcsin x = \frac{1}{\sqrt{1 - x^2}}$, but I don't see how I can use the proof backwards for integration either…

Any pointers?

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There is a standard substitution in this sort of situation, namely $x=\sin\theta$, where we assume $-\pi/2 \le \theta \le \pi/2$. Then $dx=\cos\theta\, d\theta$, and $\sqrt{1-x^2}=\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}=\cos\theta$ since in our interval $\cos$ is non-negative.

Thus $$\int \frac{dx}{\sqrt{1-x^2}}=\int \frac{\cos\theta}{\cos\theta}d\theta=\int d\theta=\theta+C.$$

But $\theta=\arcsin x$. Now it's over.

Comment 1: Regrettably, it is commonplace in solutions not to mention $-\pi/2 \le \theta \le \pi/2$, and it is commonplace to not justify $\sqrt{\cos^2\theta}=\cos\theta$. So in an integration question, in most calculus courses, the solution would be even shorter.

Comment 2: Note how close this truly standard approach is, in this case, to the suggestion by David Speyer. The difference is that the calculus teacher would not notice. The same substitution is used in many other integrals that involve $\sqrt{1-x^2}$, and close relatives can be used for integrals that involve $\sqrt{a-bx^2}$ where $a$ and $b$ are positive.

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  • $\begingroup$ OK. But, as I understand it the rule we have to use is: $\int f(x) \operatorname d x = \int g(h(x)) \cdot h'(x) \operatorname d x = \int g(t) \operatorname d t$ with $t := h(x)$. Here $g(x) := \frac{1}{\sqrt{1 - x^2}}$ and $h(x):= \sin x$ but: $g(h(x)) \cdot h'(x) = -\frac{\cos x}{\sqrt{1 - \sin^2 x}} \not= \frac{1}{\sqrt{1 - x^2}}$ $\endgroup$
    – pascal
    Jul 6, 2011 at 19:31
  • $\begingroup$ @pascal: If you are going to use this version of substitution, then you need to put $h(x)=\arcsin x$, not $\sin x$, and you have essentially broken the directive. But the substitution in the style I did it is essentially universal in calculus. Your text will have the method, maybe under the heading $u$-substitution if you are using an American text. It is early on in methods of integration. $\endgroup$
    – André Nicolas
    Jul 6, 2011 at 19:49
  • $\begingroup$ ah yes, as usual the homework required knowledge from a future lesson. Thanks for the help (well basically directly giving the answer :D) $\endgroup$
    – pascal
    Jul 6, 2011 at 19:54
  • $\begingroup$ @pascal: Giving the answer was hard to avoid, it is a one-step calculation. Conveniently it cleared a misunderstanding about the range of substitution techniques. $\endgroup$
    – André Nicolas
    Jul 6, 2011 at 20:00
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I suspect I am going to annoy your calculus teacher by writing this but:

Suppose that you are given the problem of computing $\int f(x) dx$. A little fairy comes and whispers in your ear that the answer is $g(x)$. Then you can compute this integral in a "forward" way by making the substitution $x = g^{-1}(u)$. When you do this, $f(x) dx$ should turn into $du$, which can be integrated without difficulty.

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Try substituting $x = cos(\theta)$.

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    $\begingroup$ Not $x = \sin \theta$? ;-) $\endgroup$
    – Aryabhata
    Jul 6, 2011 at 19:16
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    $\begingroup$ NO. Never!! ;-) $\endgroup$
    – Pratik Deoghare
    Jul 6, 2011 at 19:17
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Using the substition $x=\sin t$, $dx = \cos t ~dt$, we get:

$$\int \frac{dx}{\sqrt{1-x^2}} = \int \frac{\cos t ~dt}{\sqrt{1-\sin^2 t}} = \int \frac{\cos t ~dt}{\cos t} = \int dt = t.$$

By our substition $x=\sin t$ we have $t=\arcsin x$.

Therefore

$$\int \frac{dx}{\sqrt{1-x^2}} = \arcsin x + C.$$

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    $\begingroup$ Which appears as (part of) another (already accepted) answer posted a whole year before. What is your game here? $\endgroup$
    – Did
    Jul 15, 2012 at 6:14

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