Why do we only use 1st quadrant values for Trigonometric u substitution integrals?

Question

Say we have the following integral: $$ \int_0^{\frac{1}{2}} \frac{x}{\sqrt{1-x^2}} dx$$ The substitution of $x=\sin(\theta)$ would mean the limits must be changed, but when $x=1/2$, $\sin(\theta)$ must equal $\frac{1}{2}$. But there are two values of theta for $\frac{1}{2}$ - being $\frac{\pi}{6}$ and $\frac{5\pi}{6}$. Why can't you use $\frac{5\pi}{6}$? Is this not a valid solution?

Thank you in advance!

Answer 1

This is because you would be counting the area of an intermediate range of values, which was not supposed to be taken. Think about it, if you choose limits to be $0$ and $\frac {5\pi}{6}$, you would be counting intermediate values such as $\frac {\pi}{2}$. But this value was never supposed to be included, as here $x=1$. The key is to build a bijective mapping between the two variables, $x$ and $\theta$.

For example, instead of $0$ and $\frac {5\pi}{6}$, you could have chosen the limits to be $2\pi$ and $\frac {13\pi}{6}$.

Answer 2

Integration by substitution is slightly more nuanced than how it is typically conveyed in calculations-oriented calculus settings; in general, using a substitution function that is injective on the interval of integration (even though this is not strictly necessary) ensures that the substitution is valid.

Here, if your substitution function (using wikipedia’s notation) is $ϕ(x)=\arcsin(x),$ then it is indeed injective; however, since its range is $[-\frac\pi2,\frac\pi2],$ $ϕ(\frac12)=\frac\pi6\neq\frac{5\pi}6.$

Answer 3

Technically you can use $\frac{5\pi}{6}$ as one of the bounds of $\theta,$ but you have to beware of several pitfalls.

One question is about the substitution. Just looking at the antiderivative (not the bounds) for the moment, the immediate result of the substitution $x = \sin\theta$ is $$ \int \frac{\sin \theta}{\sqrt{1 - \sin^2\theta}}\, \cos\theta \,\mathrm d\theta, $$ which simplifies to $$ \int \frac{\sin \theta}{\sqrt{\cos^2\theta}}\, \cos\theta \, \mathrm d\theta. $$

We usually further simplify this to $$ \int \sin \theta \, \mathrm d\theta, $$ but that simplification assumes that $\sqrt{\cos^2\theta} = \cos\theta,$ which is true only if $\cos\theta \geq 0,$ which is not true when $\frac\pi2 < \theta < \frac{3\pi}{2}.$ You can still bound the integral by values of $\theta$ that are not in the interval $\left[-\frac\pi2,\frac\pi2\right],$ as explained in this answer, but you have to be careful that you are actually using the correct substitution. A substitution that works for all $\theta$ is $$ \int \frac{x}{\sqrt{1-x^2}}\,\mathrm dx = \int \sin \theta \operatorname{sgn}(\cos\theta)\, \mathrm d\theta, $$ where $$ \operatorname{sgn}(y) = \begin{cases}1 & y > 0,\\0 & y = 0,\\-1 & y < 0.\end{cases} $$

For example, the following is a valid substitution: \begin{align} \int_0^{1/2} \frac{x}{\sqrt{1-x^2}}\,\mathrm dx &= \int_\pi^{5\pi/6} \sin \theta \operatorname{sgn}(\cos\theta)\, \mathrm d\theta\\ &= \int_\pi^{5\pi/6} -\sin \theta \, \mathrm d\theta & \cos\theta < 0\\ &= \bigg[ \cos \theta \bigg]_\pi^{5\pi/6}\\ &= \left(-\frac{\sqrt3}2\right) - (-1)\\ &= 1 - \frac{\sqrt3}2, \end{align} which is the correct answer.

Note that when $-\pi \leq \theta \leq \frac{5\pi}{6}$ we have $0 \leq x \leq \frac12,$ and the substitution $x = \sin\theta$ really does relate each value of $\theta$ to one value of $x$ in the original interval of integration. If you instead integrate from $0$ to $\frac{5\pi}{6}$ things start to get a little weird: as $\theta$ goes from $0$ to $\frac{5\pi}{6}$, $x$ goes from $0$ to $1$ and then back down to $\frac12.$ If you're careful, the integral over $\theta$ from $\frac\pi2$ to $\frac{5\pi}{6}$ will exactly cancel the integral from $\frac{\pi}{6}$ to $\frac\pi2$, and you'll end up with exactly what you would have had if you had just integrated over $\theta$ from $0$ to $\frac{\pi}{6}$ in the first place. But you do need to be careful, since $\cos\theta$ changes sign over the interval of integration.