Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_0^1\sqrt{\frac{1-x}{1+x}}dx$ is

Unlocking the Integral

Welcome, future engineer! Today, we are going to peel back the layers of a seemingly intimidating integral. When you first look at the expression $I={\int }_0^1\sqrt{\frac{1-x}{1+x}}dx$, it is natural to feel a moment of hesitation.

It looks like a classic trap—a square root of a rational function that doesn't immediately scream 'standard formula.' But here is the secret: in JEE Advanced, the most difficult-looking problems often yield to the most elegant algebraic manipulations. Let us embark on this journey together.

The Rationalization Strategy

Imagine you are standing before a locked door. You have the key, but you are trying to use it in the wrong lock. Many students see this integral and immediately think of trigonometric substitution, like $x=\cos (2\theta )$.

While that works, it is like taking the long way around the mountain. Instead, let us look at the integrand: $\sqrt{\frac{1-x}{1+x}}$.

What if we could make the numerator a perfect square? If we multiply the numerator and the denominator by $(1-x)$, the numerator becomes $(1-x{)}^2$. The denominator becomes $(1+x)(1-x)$, which is the difference of squares: $1-x^2$.

Now, our integral looks like this:

This is the 'Aha!' moment. By pulling the $(1-x{)}^2$ out of the square root, we get $\frac{1-x}{\sqrt{1-x^2}}$. We have successfully cleared the radical from the numerator.

Remember, because our limits are $x\in [0,1]$, the term $(1-x)$ is always positive, so we don't need to worry about absolute value signs. The expression is now clean, manageable, and ready for the next step.

The Power of Linearity

Now that we have $I={\int }_0^1\frac{1-x}{\sqrt{1-x^2}}dx$, we face a fraction. But look at the numerator: it is a simple subtraction.

Integration is a linear operator, which means we can split this into two separate, simpler integrals:

Let us call these $I_1$ and $I_2$. By separating them, we have turned one 'scary' problem into two 'standard' problems. This is the strategy of a champion: divide and conquer.

Solving the Standard Forms

First, let us tackle $I_1={\int }_0^1\frac{1}{\sqrt{1-x^2}}dx$. This is a fundamental integral from your calculus toolkit. You should recognize this instantly as the derivative of ${\sin }^{-1}(x)$.

That was quick! Now, for $I_2={\int }_0^1\frac{x}{\sqrt{1-x^2}}dx$. Here, we see a function $1-x^2$ and its derivative (roughly) $x$ in the numerator.

This is a classic invitation for substitution. Let $u=1-x^2$. Then $du=-2xdx$, which means $xdx=-\frac{1}{2}du$.

Substituting this into our integral, we get:

Integrating $u^{-\frac{1}{2}}$ gives $2u^{\frac{1}{2}}$, so the constants cancel out beautifully, leaving us with $-\sqrt{u}$. Substituting back $u=1-x^2$, we get the anti-derivative $-\sqrt{1-x^2}$.

Evaluating this from $0$ to $1$:

We have arrived at the finish line. We found $I_1=\frac{\pi }{2}$ and $I_2=1$. Recalling our split, $I=I_1-I_2$, we simply subtract the two results:

Look at that result. It is elegant, precise, and perfectly matches our options. You didn't just solve a problem; you navigated a logical path, utilized the properties of integration, and executed the algebra with precision.

This is the mindset that will carry you through the JEE Advanced. Keep practicing, keep questioning, and keep falling in love with the process!