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I have a question to calculate the indefinite integral: $$\int \sqrt{1-x^2} dx $$ using trigonometric substitution.

Using the substitution $ u=\sin x $ and $du =\cos x\,dx $, the integral becomes: $$\int \sqrt{\cos^2 u} \, \cos u \,du = \int \|{\cos u}\| \cos u\, du $$

Q: (part a) At what point (if at all) is it safe to say that this is the equivalent of ? $$\int \cos^2 u\, du = \int \frac {1 + \cos 2u} {2} du$$ (this is easy to solve, btw).

In lectures, it was made abundantly clear that over certain intervals (eg $ 0 \le u \le \pi/2$) that $cos u$ is +ve and is safe to do so, but in the indefinite form, the same argument cannot be made (eg $ \pi/2 \le u \le n\pi$).

Q: (part b) Is it safe to declare it ok due to the nature of the original integral, which, using a sqrt() must return a +ve number? It could then be argued that it was the substitution which artificially added a -ve aspect...

Any suggestions on how to proceed?

PS: This is a 1st year calculus course and am revising for exams ;)

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    $\begingroup$ The original integrand allows for $x$ to be in the interval $[-1,1]$ which means that $\cos$ can be negative. In my experience ignoring this is almost never a problem. Rather than worrying about the details of when $\cos$ is positive or negative, recall that you can always prove that your final answer is correct by showing that its derivative equals the original integrand. $\endgroup$ – Spencer Jun 22 '14 at 5:21
  • $\begingroup$ The lecturer has been pounding into us the need to think correctly on fundamental concerns, which is why I raise it here. Although I understand I can show equivalency by derivation, it is not a proof (as far as I am aware) $\endgroup$ – cmroanirgo Jun 22 '14 at 5:37
  • $\begingroup$ The definition of an indefinite integral is that it is the anti-derivative of the integrand (up to an additive constant). There is no more legitimate proof. $\endgroup$ – Spencer Jun 22 '14 at 5:39
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    $\begingroup$ You have it backwards. To do this with trig substitution, $x=\sin u$, not $u=\sin x$. $\endgroup$ – alex.jordan Jun 22 '14 at 5:40
  • $\begingroup$ @Spencer I don't think that's right. $x$ is in $[-1,1]$, and since $x=\sin(u)$, well, see my answer below. $\endgroup$ – alex.jordan Jun 22 '14 at 5:44
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Since $x$ ranges from $-1$ to $1$, and you are using the substitution $x=\sin(u)$, you can make this substitution with $u\in[-\pi/2,\pi/2]$, and then $\cos(u)$ is unambiguously positive.

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  • $\begingroup$ It was the logic of discussion which had me confused. This clarifies the why. Recognising the limiting domain (to real numbers) from -1 to 1 is the key ingredient. $\endgroup$ – cmroanirgo Jun 22 '14 at 6:13
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You have received good answer which all conclude that $\sqrt{\cos^2 x} = \cos x$.

But, for the time being, let us assume you still ignore if it is safe or not. So, let us write $$\int \sqrt{1-x^2} dx=\pm \int \cos^2 u\, du =\pm \int \frac {1 + \cos 2u} {2} du=\pm \Big(\frac{u}{2}+\frac{1}{4} \sin (2 u)\Big)$$ But now, let us consider the definite integral $$\int_0^1 \sqrt{1-x^2} dx=\pm \int_0^{\frac{\pi}{2}} \cos^2 u\, du =\pm \int_0^{\frac{\pi}{2}} \frac {1 + \cos 2u} {2} du=\pm \frac{\pi}{4}$$ But now, consider now the area between the curve $y=\sqrt{1-x^2}$ and the $x$ axis. All the curve is above the axis and the area is then positive. So, $\pm$ should be just replaced by $+$.

Is this making things clearer ?

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Note that there is a difference between substitution as one initially learns it (let $u=g(x)$ and what in the OP is called "trigonometric substitution." The latter process, when one is feeling pedantic, is actually called inverse trigonometric substitution.

In the example we are discussing, the substitution "really is" let $u=\arcsin x$. So $u$ naturally lives in the interval $[-\pi/2,\pi/2]$, and therefore $\cos u$ is non-negative.

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Hint: first integrate by parts to get,

$$\int\sqrt{1-x^2}\,\mathrm{d}x=x\sqrt{1-x^2}+\int\frac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x.$$

Now attempt the trigonometric substitution.

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To answer your question: yes, it is safe and doesn't really matter. When you do trig. substitution in your first year calculus course, you are always assuming that $\cos$ is positive as a result you can do:

$$\sqrt{\cos^2 x} = \cos x$$

and not have any problems. Also, take into account what @spencer said; whatever your final answer, you can just find it's derivative and prove yourself right or wrong.

Good luck on your exams!

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  • $\begingroup$ We have spent a lot of time on your statement. It is not true at 3*pi/2. I can personally guarantee, that in my course, making that assumption without qualification means loss of marks. $\endgroup$ – cmroanirgo Jun 22 '14 at 5:38
  • $\begingroup$ Absolutely right but for first year, as far as I know, the integral is calculated with keeping in mind that $\cos$ is in between that intervals which you specified, otherwise $\sqrt{\cos^2 x} \rightarrow \cos x$ doesn't make sense. Depending on which textbook you use, go to the trig. substitution chapter and from an example with $\cos x$ it will say specifically that $\cos x$ is bounded between $0 \le x \le \pi/2$ $\endgroup$ – Jeel Shah Jun 22 '14 at 5:40
  • $\begingroup$ PS; This is an indefinite integral. No boundaries are specified $\endgroup$ – cmroanirgo Jun 22 '14 at 5:41
  • $\begingroup$ The bounds for the integral and the bounds for the function are not the same. I believe that is where you are confused. All we are saying is that we want to consider $\cos x$ when it is positive and therefore, bound it between those bounds. This doesn't imply that the entire integral is bounded between those. $\endgroup$ – Jeel Shah Jun 22 '14 at 5:42

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