Animated Solution for Mathematics - Definite Integration: The area of the region bounded by the curves $x(1+y^2)=1$ and $y^2=2x$ is:

Welcome, fellow explorer of the mathematical landscape! Today, we are tackling a classic JEE Advanced problem that tests not just your integration skills, but your ability to choose the most elegant path.

We are looking for the area bounded by two curves: $x(1+y^2)=1$ and $y^2=2x$.

At first glance, these might look like standard functions, but let's observe their structure. The first curve, $x=\frac{1}{1+y^2}$, is a variation of the 'Witch of Agnesi,' while the second, $x=\frac{y^2}{2}$, is our trusty parabola.

The key to this problem is realizing that both are functions of $y$. By shifting our perspective to integrate with respect to $y$, we transform a potentially messy problem into a beautiful, straightforward calculation.

In many calculus problems, we are conditioned to think in terms of $y=f(x)$. But here, the equations practically beg us to use $x=g(y)$.

Let's rewrite them: the first becomes $x=\frac{1}{1+y^2}$ and the second becomes $x=\frac{y^2}{2}$. When we visualize the region, we see that the area is trapped between these two curves.

To find the boundaries of this trap, we must find where they meet. We set them equal:

Now, we face the algebra. Cross-multiplying gives us $y^2(1+y^2)=2$, which expands to $y^4+y^2-2=0$.

This looks like a quadratic in disguise! Let $u=y^2$. Then $u^2+u-2=0$.

Factoring this, we get $(u+2)(u-1)=0$. Since $u=y^2$ must be non-negative for real $y$, we discard $y^2=-2$.

We are left with $y^2=1$, which gives us our limits: $y=1$ and $y=-1$. These are the vertical boundaries of our integration.

Here is where we save precious time. Notice that both $x=\frac{1}{1+y^2}$ and $x=\frac{y^2}{2}$ are even functions of $y$.

This means the entire region is perfectly symmetric about the $x$-axis. Instead of calculating the integral from $-1$ to $1$, we can calculate the area from $0$ to $1$ and simply double it.

This is a classic JEE trick—always look for symmetry to simplify your work! Our integral becomes:

Now, we integrate term by term. The integral of $\frac{1}{1+y^2}$ is the standard ${\tan }^{-1}(y)$, and the integral of $\frac{y^2}{2}$ is $\frac{y^3}{6}$.

Applying the limits from $0$ to $1$, we get:

Substituting the upper limit $y=1$, we get ${\tan }^{-1}(1)-\frac{1}{6}=\frac{\pi }{4}-\frac{1}{6}$. Substituting the lower limit $y=0$, we get ${\tan }^{-1}(0)-0=0$.

Finally, multiplying by $2$, we arrive at:

And there it is! The final area is $\frac{\pi }{2}-\frac{1}{3}$. Remember, the beauty of these problems lies in the setup. By choosing the right variable and leveraging symmetry, you turn a daunting task into a moment of clarity.