Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_0^{1}x{\cot }^{-1}(1-x^2+x^4)dx$ is :-

The integral provided is $I={\int }_0^{1}x{\cot }^{-1}(1-x^2+x^4)dx$. At first glance, the quartic polynomial inside the inverse cotangent function appears intimidating.

However, in the context of JEE Advanced, such complexity is often a mask for a hidden, elegant structure. Our goal is to peel back this mask using trigonometric identities and algebraic manipulation.

Working with ${\cot }^{-1}$ is rarely the most efficient path in calculus. We recall the identity ${\cot }^{-1}(z)={\tan }^{-1}(\frac{1}{z})$ for $z>0$.

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

Next, we perform algebraic surgery on the denominator. We rewrite $1-x^2+x^4$ as $1+x^2(x^2-1)$.

This creates a structure resembling $1+AB$, where $A=x^2$ and $B=x^2-1$. Note that $A-B=x^2-(x^2-1)=1$.

Using the identity ${\tan }^{-1}(A)-{\tan }^{-1}(B)={\tan }^{-1}\left(\frac{A-B}{1+AB}\right)$, our integral transforms into:

With the expression simplified, we use the substitution $t=x^2$, which implies $dt=2xdx$, or $xdx=\frac{1}{2}dt$. The limits remain $0$ to $1$.

The integral becomes:

We split this into two parts:

We now invoke the King's Property: ${\int }_0^{a}f(t)dt={\int }_0^{a}f(a-t)dt$. Applying this to the second integral, we replace $t$ with $1-t$:

Substituting this back, the two negative signs cancel out. We are left with:

We have reduced the quartic integral to the simple task of integrating ${\tan }^{-1}(t)$. We use integration by parts, where $u={\tan }^{-1}(t)$ and $dv=dt$.

The formula $\int udv=uv-\int vdu$ yields:

The integral of $\frac{t}{1+t^2}$ is $\frac{1}{2}\ln (1+t^2)$. Thus, our result is:

Evaluating at the limits, we obtain:

The final result is: