Animated Solution for Mathematics - Indefinite Integration: Let $f(x)=\int \frac{dx}{(3+4x^2)\sqrt{4-3x^2}},|x|<\frac{2}{\sqrt{3}}$. If $f(0)=0$ and $f(1)=\frac{1}{\alpha \beta }{\tan }^{-1}\left(\frac{\alpha }{\beta }\right),\alpha ,\beta >0$, then ${\alpha }^2+{\beta }^2$ is equal to _______.

Welcome, warrior of the JEE. Today, we face an integral that might seem daunting at first glance:

When you see a quadratic expression outside a square root and another quadratic inside, your intuition should immediately scream substitution. This is not a random problem; it is a carefully constructed puzzle designed to test your ability to simplify complex radicals.

The key to unlocking this is the substitution $x=\frac{1}{t}$. By setting $x=\frac{1}{t}$, we find that $dx=-\frac{1}{t^2}dt$.

This transformation is like finding the master key to a locked door. When we substitute this into our integral, the $t$ terms in the denominator will cancel out with the $t^2$ from the differential, leaving us with a much cleaner expression:

Now that we have simplified the integral, we are left with $\sqrt{4t^2-3}$. This radical is still a nuisance.

To eliminate it completely, we perform a second, more surgical substitution. Let the entire expression inside the root be ${\lambda }^2$, so ${\lambda }^2=4t^2-3$.

Differentiating both sides gives $2\lambda d\lambda =8tdt$, or $tdt=\frac{\lambda }{4}d\lambda$. This is where the magic happens.

By substituting this into our integral, the $\lambda$ in the numerator and the $\lambda$ from the square root cancel each other out perfectly. We are left with the standard integral:

This is the moment of triumph where the complexity collapses into a simple, solvable form.

To solve $-\int \frac{d\lambda }{3{\lambda }^2+25}$, we factor out the $3$ to get:

This is the classic form $\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$. Applying this, we get:

Now, we must return to our original variable $x$. Substituting $\lambda =\frac{\sqrt{4-3x^2}}{x}$, we obtain our function $f(x)$.

But wait—we have a constant $C$ to determine. The problem states $f(0)=0$.

As $x\to 0^{+}$, the argument of the ${\tan }^{-1}$ function approaches infinity, and ${\tan }^{-1}(\infty )=\frac{\pi }{2}$. Thus, $0=-\frac{\sqrt{3}}{15}\left(\frac{\pi }{2}\right)+C$, which gives $C=\frac{\sqrt{3}\pi }{30}$.

Finally, we evaluate $f(1)$. Substituting $x=1$ into our expression, we get:

Factoring out $\frac{\sqrt{3}}{15}$, we are left with $\frac{\sqrt{3}}{15}\left(\frac{\pi }{2}-{\tan }^{-1}\left(\frac{\sqrt{3}}{5}\right)\right)$.

Using the identity $\frac{\pi }{2}-{\tan }^{-1}(z)={\cot }^{-1}(z)$, this becomes $\frac{\sqrt{3}}{15}{\cot }^{-1}\left(\frac{\sqrt{3}}{5}\right)$, which is equivalent to $\frac{1}{5\sqrt{3}}{\tan }^{-1}\left(\frac{5}{\sqrt{3}}\right)$.

Comparing this with $\frac{1}{\alpha \beta }{\tan }^{-1}\left(\frac{\alpha }{\beta }\right)$, we find $\alpha =5$ and $\beta =\sqrt{3}$.

The final answer, ${\alpha }^2+{\beta }^2=25+3=28$. You have conquered the integral!