Animated Solution for Mathematics - Definite Integration: Comprehension Passage

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are uncovering the hidden elegance of calculus.

When you first look at an integral like $S_1={\int }_{\frac{\pi }{8}}^{\frac{3\pi }{8}}{\sin }^{2}xdx$, it is easy to feel overwhelmed. You might wonder if you need complex trigonometric identities or substitutions, but the key lies in the limits: $\frac{\pi }{8}$ and $\frac{3\pi }{8}$.

Whenever you see symmetric limits in a definite integral, your intuition should immediately invoke the King's Rule. This property states:

In our case, the sum of the limits is $a+b=\frac{\pi }{8}+\frac{3\pi }{8}=\frac{4\pi }{8}=\frac{\pi }{2}$. This is the key that unlocks the door.

When we replace $x$ with $\frac{\pi }{2}-x$, our integrand ${\sin }^{2}x$ transforms into ${\sin }^2(\frac{\pi }{2}-x)$. Because $\sin (\frac{\pi }{2}-x)=\cos x$, our integral becomes:

We now have two equivalent expressions for $S_1$: 1. $S_1={\int }_{\frac{\pi }{8}}^{\frac{3\pi }{8}}{\sin }^{2}xdx$ 2. $S_1={\int }_{\frac{\pi }{8}}^{\frac{3\pi }{8}}{\cos }^{2}xdx$

Adding these together, we obtain:

The complexity vanishes, leaving a simple linear evaluation:

Thus, $S_1=\frac{\pi }{8}$, and our target value $\frac{16S_1}{\pi }$ yields exactly 2.

Now, let us turn our attention to $S_2={\int }_{\frac{\pi }{8}}^{\frac{3\pi }{8}}{\sin }^{2}x\cdot |4x-\pi |dx$. At first glance, the modulus $|4x-\pi |$ looks like a barrier.

However, applying the substitution $x\to \frac{\pi }{2}-x$ reveals that the modulus term remains invariant:

Adding the two versions of $S_2$ yields:

The function $y=|4x-\pi |$ describes a V-shaped graph with its vertex at $x=\frac{\pi }{4}$. Between our limits of $\frac{\pi }{8}$ and $\frac{3\pi }{8}$, this graph forms two identical right-angled triangles.

The base of each triangle is the distance from the vertex to the limit: $\frac{\pi }{4}-\frac{\pi }{8}=\frac{\pi }{8}$. The height at the endpoint $x=\frac{\pi }{8}$ is $|4(\frac{\pi }{8})-\pi |=|\frac{\pi }{2}-\pi |=\frac{\pi }{2}$.

Using the area formula $\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}$, the area of one triangle is:

Since we have two such triangles, the total integral $2S_2$ is $2\times \frac{{\pi }^2}{32}=\frac{{\pi }^2}{16}$. This gives $S_2=\frac{{\pi }^2}{32}$.

Finally, our target $\frac{48S_2}{{\pi }^2}$ becomes: