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Assume we evaluate $\int \sqrt{4-x^2}\,dx $ by using the substitution $x=2 \sin(t)$ from which it follows that $4-x^2 = 4\cos^2 (t) $ so that $\int \sqrt{4-x^2}\,dx =4\int \sqrt{\cos^2 (t)} \cos(t)\,dt $.

In order to get $\int \sqrt{4-x^2}\,dx = 2\arcsin(x/2)+\frac{x\sqrt{4-x^2}}{2}+C$, we need to write $\sqrt{\cos^2 (t)} = \cos(t)$ .

My question is -

what are the arguments that allows us to assume that $\cos(t)\geq 0 $ so that $\sqrt{\cos^2 (t)} = \cos(t)$?

(taking $\sqrt{\cos^2 (t)} = -\cos(t)$ for example yields a wrong result)

Bernard
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Dr. John
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  • $\sqrt{x}$ always means the positive square root of $x$ by convention. This has been mentioned in many previous questions, such as this one. – Toby Mak Dec 21 '19 at 14:26
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    @TobyMak that's not the point of the question, I think. He knows that $\sqrt{\cos^2 x}$ = $\lvert \cos x \rvert$, but if $\cos x < 0$, that's $-\cos x$ (which makes the integration go wrong) – Izaak van Dongen Dec 21 '19 at 14:28
  • Because the sine function produces all of its values on the interval $[-\frac \pi 2,\frac \pi 2]$, you may assume $t\in [-\frac \pi 2,\frac \pi 2]$. This ensures that $\cos t\geq 0$. – bjorn93 Dec 21 '19 at 14:31
  • By the way, you can also integrate this by parts without a trig. substitution. – bjorn93 Dec 21 '19 at 15:01

1 Answers1

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The range of definition of the integrand is $-2\leq x\leq2$, so $-1\leq\sin t\leq1$. Therefore $-\frac{\pi}{2}\leq t\leq\frac{\pi}{2}$ (replacement must be monotonic) and $\cos t\geq0$.

Bernard
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thing
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  • Got it. Thanks! – Dr. John Dec 21 '19 at 14:35
  • @Dr.John, I’ll add that this is also related to the range of the arcsine function when doing the reverse change. – thing Dec 21 '19 at 14:38
  • Dear @thing - many thanks! I absolutely agree. The problem is that in contrast to the multivariable case, teachers often forget to mention that the replacement should be injective also in the single variable case. The $arcsin$ argument indeed provides an interesting further insight. Thanks! – Dr. John Dec 21 '19 at 14:43