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Problems that deceive you #1

Let us look at one of the most easiest problems in calculus
$$\int_0^{\frac{\pi}{2}}\sqrt{4\sin^2\frac{x}{2}-4\sin\frac{x}{2}+1}\; dx$$
Seems pretty simple and straightforward, doesn't it? Inside the square root, the expression evaluates to $(2\sin\frac{x}{2}-1)^2$, and thus the integral will be
$$\int_0^{\frac{\pi}{2}}\Bigl(2\sin\frac{x}{2}-1\Bigr)dx$$
This then can be evaluated by the following:
$$\biggl[-4\cos\frac{x}{2} -x\biggr]_0^{\frac{\pi}{2}}$$
Hold on for the surprise...

Its wrong!


What seemed like a really straightforward and easy problem, is designed in such a way to appear easy so that we approach the wrong way.

Why is the approach wrong?

We did not consider the fact that a simple square root symbol represents its principal square root i.e., the positive square root. That means each and every value the expression inside the original integral evaluates to a positive quantity. But that is not the case when we simplified the integral to 
$$\int_0^{\frac{\pi}{2}}\Bigl(2\sin\frac{x}{2}-1\Bigr)dx$$
In the above integral, the expression inside can also evaluate to a negative quantity, precisely when $\sin \frac{x}{2} \lt \frac{1}{2}$. So we must find an another way out of this.

The correct approach

The original integral evaluates to 
$$\int_0^{\frac{\pi}{2}}\sqrt{\Bigl(2\sin\frac{x}{2}-1\Bigr)^2}\; dx$$
This can be split as 
$$\int_0^{\frac{\pi}{6}}\sqrt{\Bigl(2\sin\frac{x}{2}-1\Bigr)^2}+\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\sqrt{\Bigl(2\sin\frac{x}{2}-1\Bigr)^2}\; dx$$
When we are cancelling out the square roots, we must be sure to make the outside expression positive. We all know that
$$1-2\sin\frac{x}{2}\ge 0, x\in \Bigl[0, \frac{\pi}{6}\Bigr]$$
$$2\sin\frac{x}{2}-1\ge 0, x\in \Bigl[\frac{\pi}{6}, \frac{\pi}{2}\Bigr]$$
Applying the above formulae in the split integral, we obtain
$$\int_0^{\frac{\pi}{6}}\Bigl(1-2\sin\frac{x}{2}\Bigr)+\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\Bigl(2\sin\frac{x}{2}-1\Bigr)\; dx$$
which then can be evaluated into a value.
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