Confusion in bounds while substituting $x=\sin^2\theta$ in $\int_{0}^{1}\frac{x^{2}}{\sqrt{x(1-x)}}\mathrm dx$

Question

I really need help in evaluating the following integral from MIT-BEE Quarterfinals $2025$.

The question is to evaluate the integral $$\int_{0}^{1}\frac{x^{2}}{\sqrt{x(1-x)}}\mathrm dx$$

Now I evaluated this given integral in the following way :

I substituted $x=\sin^{2}(\theta)$ and found the lower and the upper limits to be $0$ and $\frac{\pi}{2}$.

The integral becomes

$$2\int_{0}^{\frac{\pi}{2}}\frac{\sin^{4}(\theta)\times\sin(\theta)\times\cos(\theta)}{\sqrt{\sin^{2}(\theta)\times\cos^{2}(\theta)}}\mathrm d\theta$$

which gives the result as

$$2\int_{0}^{\frac{\pi}{2}}\sin^{4}(\theta)\mathrm d\theta$$

because $\sqrt{\sin^{2}(\theta)×\cos^{2}(\theta)}$ is equal to $[\sin(\theta)×\cos(\theta)]$ and it will cancel out in both the numerator and denominator.

Now, I evaluated $2\int_{0}^{\frac{\pi}{2}}\sin^{4}(\theta)\mathrm d\theta$ very easily.

$$\begin{align}2\int_{0}^{\frac{\pi}{2}}\sin^{4}(\theta)\mathrm d\theta&=2\int_{0}^{\frac{\pi}{2}}\sin^{2}\theta(1-\cos^{2}\theta)\mathrm d\theta\\&=2\int_{0}^{\frac{\pi}{2}}\sin^{2}\theta\mathrm d\theta-2\int_{0}^{\frac{\pi}{2}}\sin^{2}\theta×\cos^{2}\theta\,\mathrm d\theta\end{align}$$

Now, the integral became very easy after this step.

But I am having a small doubt in this process.

When I am substituting $$x=\sin^{2}\theta$$ I am getting $2$ values of $\theta$ for $$\sin^{2}\theta=1$$ which are $+\frac{\pi}{2}$ and $-\frac{\pi}{2}$.

But why should I consider only $+\frac{\pi}{2}$ and neglect $-\frac{\pi}{2}$?

I discussed this integral with many of my lecturers, but they are also unable to guide me in my doubt.

Please help me out with the logic.

Answer 1

I don't see why we can't consider $-\pi/2$. Then our integral becomes $$ 2\int_0^{-\pi/2} \frac{\sin^4\theta\sin\theta\cos\theta}{\sqrt{\sin^2\theta\cos^2\theta}} = -2\int_{-\pi/2}^0 \frac{\sin^4\theta\sin\theta\cos\theta}{\sqrt{\sin^2\theta\cos^2\theta}}. $$When $\theta$ ranges from $-\pi/2$ to $0$, we have that $\sin\theta$ is negative. However, when we square $\sin$ and take a square root, it becomes positive. Thus our integral becomes $$ 2\int_{-\pi/2}^0 \sin^4\theta. $$And it shouldn't be hard to convince yourself this value is the same as your integral.

Answer 2

Note that whenever we make a substitution to evaluate an integral, it must be bijective so that back-substitution can be done. Your substitution $x=\sin^2\theta$ can be broken down into $2$ cases as follows so that it is bijective:

• $\sqrt x=\sin\theta$, then the domain of $\theta$ can be $\left[2n\pi,2n\pi+\frac\pi2\right]$ or $\left[2n\pi+\frac\pi2,(2n+1)\pi\right]$ for $n\in\mathbb Z$. A convenient choice can be $\left[0,\frac\pi2\right]$
• $-\sqrt x=\sin\theta$, then the domain of $\theta$ can be $\left[2n\pi-\frac\pi2,2n\pi\right]$ or $\left[(2n+1)\pi,(2n+1)\pi+\frac\pi2\right]$ for $n\in\mathbb Z$. A convenient choice can be $\left[-\frac\pi2,0\right]$.

So that's why if $0$ is the lower bound, the upper bound has to be $\frac\pi2$. If you want to involve $0$ and $-\frac\pi2$, the bounds would be $\displaystyle\int_{-\frac\pi2}^0\cdots\mathrm d\theta$.