Evaluate $\int \sqrt{1-x^2}\,dx$

Question

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 ;)

Answer 1

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.

Answer 2

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.

Answer 3

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!

Answer 4

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 ?

Answer 5

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.