Animated Solution for Mathematics - Indefinite Integration: Evaluate the following $\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$

Welcome, student. Take a deep breath. I know exactly what you are feeling when you look at the integral $I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$. It looks like a tangled mess of nested roots, a classic 'scare tactic' problem designed to test your composure under pressure.

But here is the secret of JEE Advanced: complexity is often just a mask for a very elegant, simple truth. Our job today is to peel back that mask.

The Geometric Intuition

When you see an expression of the form $\frac{1-u}{1+u}$, your mathematical intuition should immediately fire. This is not just a random fraction; it is the signature of the half-angle identities. We are looking at a structure that begs to be converted into trigonometry.

We choose the substitution $x={\cos }^2\theta$. Why? Because it forces $\sqrt{x}=\cos \theta$. Suddenly, the integrand becomes $\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}$.

The radical is no longer a monster; it is a gateway to simplification. If we change $x$, we must change the differential $dx$. Differentiating $x={\cos }^2\theta$ gives us:

This negative sign is a common trap—do not lose it! It is the ghost in the machine that will haunt your final answer if you aren't careful.

The Collapse of the Radical

Now, let us bring these pieces together. Our integral transforms into:

Recall your half-angle identities: $1-\cos \theta =2{\sin }^2(\frac{\theta }{2})$ and $1+\cos \theta =2{\cos }^2(\frac{\theta }{2})$. When you plug these into the radical, the $2$s cancel out, and you are left with $\sqrt{{\tan }^2(\frac{\theta }{2})}$, which is simply $\tan (\frac{\theta }{2})$.

We have traded an algebraic nightmare for a trigonometric expression:

The Algebraic Grind

We are not done yet. We have a $\tan (\frac{\theta }{2})$ and a $\sin \theta$. To integrate this, we need to align our angles.

We expand $\tan (\frac{\theta }{2})$ as $\frac{\sin (\frac{\theta }{2})}{\cos (\frac{\theta }{2})}$ and use the double-angle identity $\sin \theta =2\sin (\frac{\theta }{2})\cos (\frac{\theta }{2})$.

Watch the cancellation: the $\cos (\frac{\theta }{2})$ in the denominator of the tangent term cancels perfectly with the one in the sine expansion. We are left with:

The Final Stretch

We use the identity $2{\sin }^2(\frac{\theta }{2})=1-\cos \theta$ to linearize our expression. Our integral becomes:

Integrating $\cos \theta$ is trivial. For ${\cos }^2\theta$, we use the power-reduction identity ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$.

After performing the integration and simplifying, we arrive at:

The Return to Reality

Finally, we must return to the world of $x$. We know $x={\cos }^2\theta$, so $\theta ={\cos }^{-1}\sqrt{x}$. Using our right-triangle visualization, where the adjacent side is $\sqrt{x}$ and the hypotenuse is $1$, we find the opposite side is $\sqrt{1-x}$.

Thus, $\sin \theta =\sqrt{1-x}$. Substituting these back, we get our final, elegant result:

Take a moment to appreciate this. You started with a terrifying square root, and through the systematic application of identities and substitutions, you tamed it. This is the essence of JEE Advanced mathematics—not brute force, but the elegant application of logic.