Animated Solution for Mathematics - Definite Integration: The value of ${\int }_0^{\pi }|\cos x{|}^{3}dx$

The problem asks us to evaluate the integral $I={\int }_0^{\pi }|\cos x{|}^{3}dx$. At first glance, the modulus sign $|\dots |$ acts as a gatekeeper, ensuring the function remains non-negative.

If you ignore the modulus, you will calculate the net area rather than the total area. We must respect the geometric reality that the function is strictly non-negative.

Imagine the graph of $\cos x$. It starts at $1$ at $x=0$, drops to $0$ at $x=\frac{\pi }{2}$, and dives into negative territory, reaching $-1$ at $x=\pi$.

When we cube the function, the negative values remain negative. However, the modulus $|\cos x{|}^3$ reflects that negative section back above the $x$-axis.

Because of this behavior, we cannot integrate from $0$ to $\pi$ in one go. We must split the integral at the point where the function changes sign: $x=\frac{\pi }{2}$.

By splitting the integral, we adhere to the definition of the modulus. In the interval $[0,\frac{\pi }{2}]$, $\cos x$ is positive, so $|\cos x{|}^3={\cos }^{3}x$.

In the interval $(\frac{\pi }{2},\pi ]$, $\cos x$ is negative, so $|\cos x{|}^3=(-\cos x{)}^3=-{\cos }^{3}x$. Our integral becomes:

This transformation turns a tricky modulus problem into two standard integrals.

To integrate ${\cos }^{3}x$, we linearize the power using the triple angle identity: $\cos 3x=4{\cos }^{3}x-3\cos x$. Rearranging this, we obtain:

This substitution is a classic JEE maneuver. It converts a complex power into simple, integrable cosine terms. Substituting this into our expression, we get:

Integrating is now straightforward. The integral of $\cos 3x$ is $\frac{\sin 3x}{3}$, and the integral of $3\cos x$ is $3\sin x$.

Applying the limits, the first part evaluates to $\frac{2}{3}$. The second part, due to the limits and the negative sign, also contributes $\frac{2}{3}$. Adding them together, we get:

The final result is $\frac{4}{3}$.

Wallis' Formula

Because the function $|\cos x{|}^3$ is symmetric about $x=\frac{\pi }{2}$, the area from $0$ to $\frac{\pi }{2}$ is exactly half the total area. We can use Wallis' Formula:

For $n=3$, this yields $\frac{2}{3}\times 1=\frac{2}{3}$. Doubling this gives us $\frac{4}{3}$ instantly. Mathematics is about finding the most elegant path; keep practicing to spot these symmetries!