Animated Solution for Mathematics - Definite Integration: If ${\int }_0^{1}4{\cot }^{-1}(1-2x+4x^2)dx=a{\tan }^{-1}(2)-b{\log }_e(5)$, where $a,b\in \mathbb{N}$, then $(2a+b)$ is equal to .........

Welcome, fellow explorers of the mathematical universe. Today, we stand before a problem that, at first glance, might seem like a daunting mountain of calculus.

We are tasked with evaluating the integral $I={\int }_0^{1}4{\cot }^{-1}(1-2x+4x^2)dx$. It looks complex, perhaps even impossible, but in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

The Reciprocal Shift

Our first instinct when facing an inverse trigonometric function like ${\cot }^{-1}$ is to seek a more familiar territory. We know that ${\cot }^{-1}(x)={\tan }^{-1}(\frac{1}{x})$ for $x>0$.

Since our limits of integration are from $0$ to $1$, the argument $1-2x+4x^2$ is always positive. Thus, we can rewrite our core integral $J$ as:

This transformation is our first victory. We have moved from the unfamiliar to the familiar.

The Algebraic Dance

Now, look closely at that denominator: $1-2x+4x^2$. It is begging to be manipulated. We want to force it into the form $1+AB$ to utilize the identity ${\tan }^{-1}(\frac{A-B}{1+AB})={\tan }^{-1}(A)-{\tan }^{-1}(B)$.

Let us rearrange the terms:

Here, we have identified our $A$ and $B$. If $A=2x$ and $B=2x-1$, then $A-B=2x-(2x-1)=1$. The numerator of our integrand is exactly $1$.

This is not a coincidence; it is the design of the problem. Our integral now becomes:

Applying the identity, we split the integral into two simpler parts:

The Vanishing Act

Let us focus on the second integral, $K={\int }_0^1{\tan }^{-1}(2x-1)dx$. This is where the 'King's Rule' shines. By replacing $x$ with $(1-x)$, we get:

Since ${\tan }^{-1}(-x)=-{\tan }^{-1}(x)$, we have $K=-{\int }_0^1{\tan }^{-1}(2x-1)dx=-K$. This implies $2K=0$, so $K=0$.

The second integral has vanished into thin air, leaving us with only the first part!

The Final Integration

We are left with $J={\int }_0^1{\tan }^{-1}(2x)dx$. We solve this using integration by parts, letting $u={\tan }^{-1}(2x)$ and $dv=dx$.

The formula $\int udv=uv-\int vdu$ gives us:

The boundary term evaluates to ${\tan }^{-1}(2)$. The remaining integral is a simple logarithmic form, $\frac{1}{4}\ln (1+4x^2)$, which evaluates to $\frac{1}{4}\ln (5)$.

Thus, $J={\tan }^{-1}(2)-\frac{1}{4}\ln (5)$.

Multiplying by the constant $4$ we set aside at the start, we get $I=4{\tan }^{-1}(2)-\ln (5)$. Comparing this to $a{\tan }^{-1}(2)-b\ln (5)$, we find $a=4$ and $b=1$.

The final value $2a+b=2(4)+1=9$. We have conquered the mountain!