Animated Solution for Mathematics - Definite Integration: Find the value of ${\int }_{-\pi /3}^{\pi /3}\frac{\pi +4x^3}{2-\cos \left(|x|+\frac{\pi }{3}\right)}dx$.

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of trigonometric functions and algebraic terms.

When you see an integral like $I={\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}\frac{\pi +4x^3}{2-\cos \left(|x|+\frac{\pi }{3}\right)}dx$, your heart might skip a beat. It looks intimidating, but in the world of JEE Advanced, intimidation is often just a mask for elegance. Let us peel back the layers.

The Power of Symmetry

Whenever you see symmetric limits of integration—in this case, from $-\frac{\pi }{3}$ to $\frac{\pi }{3}$—your brain should immediately light up. This is a massive hint that the universe of integration is inviting you to look for symmetry.

We have a numerator with two distinct parts: a constant $\pi$ and an algebraic term $4x^3$. Let us split the integral into two separate entities:

Look at that second integral. Let $f(x)=\frac{4x^3}{2-\cos (|x|+\frac{\pi }{3})}$. If we replace $x$ with $-x$, the numerator becomes $4(-x{)}^3=-4x^3$, while the denominator remains unchanged due to the modulus property $|-x|=|x|$.

Thus, $f(-x)=-f(x)$, which is the definition of an odd function. The area under an odd function over a symmetric interval is zero, meaning the negative area on the left perfectly cancels the positive area on the right. Half of our problem vanishes into thin air, leaving us with only the first part.

Simplifying the Even Function

Now we are left with $I={\int }_{-\frac{\pi }{3}}^{\frac{\pi }{3}}\frac{\pi }{2-\cos (|x|+\frac{\pi }{3})}dx$. Since the integrand is now an even function, we can use the property ${\int }_{-a}^{a}g(x)dx=2{\int }_0^{a}g(x)dx$.

This allows us to focus only on the positive domain $[0,\frac{\pi }{3}]$. In this domain, $|x|=x$, so we can drop the modulus sign entirely. Our integral becomes:

To simplify further, let $t=x+\frac{\pi }{3}$, which implies $dt=dx$. When $x=0$, $t=\frac{\pi }{3}$, and when $x=\frac{\pi }{3}$, $t=\frac{2\pi }{3}$. Our integral transforms into:

The Tangent Half-Angle Transformation

We have arrived at a standard form. Whenever you see a constant plus or minus a cosine or sine in the denominator, the tangent half-angle substitution is your best friend.

We use the identity $\cos t=\frac{1-{\tan }^2(t/2)}{1+{\tan }^2(t/2)}$. Substituting this into our integral gives us:

After taking the common denominator and simplifying, the expression becomes:

Simplifying the denominator yields $2+2{\tan }^2(t/2)-1+{\tan }^2(t/2)=1+3{\tan }^2(t/2)$. Recalling that $1+{\tan }^2(t/2)={\sec }^2(t/2)$, we recognize that ${\sec }^2(t/2)$ is the derivative of $\tan (t/2)$, perfectly setting us up for the final substitution.

The Final Integration

Let $u=\sqrt{3}\tan (t/2)$. Then $du=\frac{\sqrt{3}}{2}{\sec }^2(t/2)dt$, which means ${\sec }^2(t/2)dt=\frac{2}{\sqrt{3}}du$.

Updating our limits: when $t=\frac{\pi }{3}$, $u=\sqrt{3}\tan (\frac{\pi }{6})=1$. When $t=\frac{2\pi }{3}$, $u=\sqrt{3}\tan (\frac{\pi }{3})=3$. The integral becomes:

This is the standard integral for ${\tan }^{-1}(u)$. Evaluating this, we get:

Since ${\tan }^{-1}1=\frac{\pi }{4}$, our final answer is:

We started with a terrifying expression and, through the systematic application of symmetry, substitution, and trigonometric identities, we arrived at a precise, elegant result. This is the essence of mathematics—taking the complex and making it simple.