Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_0^{1/2}\frac{1+\sqrt{3}}{((x+1{)}^2(1-x){)}^{1/4}}dx$ is ________.

Welcome, fellow traveler on the road to JEE Advanced. Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of radicals and powers.

We are looking at the integral $I={\int }_0^{1/2}\frac{1+\sqrt{3}}{((x+1{)}^2(1-x){)}^{1/4}}dx$.

When you see an expression like this, it is natural to feel a moment of hesitation. But remember, in the world of competitive mathematics, complexity is often just a mask for elegance waiting to be revealed.

The Geometric Intuition

Our first goal is to simplify the integrand. We see the term $((x+1{)}^2(1-x){)}^{1/4}$.

The presence of $(1-x)$ and the square root structure suggests that we should look for a trigonometric identity. If we let $x=\sin \theta$, then $1-x$ becomes $1-\sin \theta$, and the differential $dx$ becomes $\cos \theta d\theta$.

By substituting $x=\sin \theta$, we transform our limits: when $x=0$, $\theta =0$, and when $x=1/2$, $\theta =\pi /6$. The integral now reads:

Because $\sqrt{{\cos }^2\theta }=\cos \theta$ (since $\theta$ is in the first quadrant), the $\cos \theta$ terms in the numerator and denominator perform a beautiful dance and cancel each other out.

We are left with the much friendlier integral:

The Power of Half-Angles

Now, we face a classic hurdle: integrating the reciprocal of $(1-\sin \theta )$. This is where the Weierstrass substitution, or the half-angle tangent substitution, becomes our best friend.

We set $y=\tan (\theta /2)$. This substitution is a heavy hitter in the JEE arsenal because it converts trigonometric functions into rational algebraic ones.

We know that $\sin \theta =\frac{2y}{1+y^2}$ and $d\theta =\frac{2dy}{1+y^2}$.

As we update our limits, we encounter $\tan (15^{\circ })$, which is $2-\sqrt{3}$. Substituting these into our integral, we get:

Watch closely as the denominator $1-\frac{2y}{1+y^2}$ simplifies to $\frac{1+y^2-2y}{1+y^2}=\frac{(1-y{)}^2}{1+y^2}$. The $(1+y^2)$ terms cancel out perfectly, leaving us with a simple power rule problem:

The Final Resolution

We are now in the home stretch. Integrating $(1-y{)}^{-2}$ with respect to $y$ gives us $\frac{1}{1-y}$.

Applying the limits from $0$ to $2-\sqrt{3}$, we calculate the difference:

Rationalizing $\frac{1}{\sqrt{3}-1}$ gives us $\frac{\sqrt{3}+1}{2}$. Subtracting $1$ yields $\frac{\sqrt{3}+1-2}{2}=\frac{\sqrt{3}-1}{2}$.

Finally, multiplying by our constant $2(1+\sqrt{3})$, we see the magic happen:

And there it is. The complexity collapses into the integer 2.

This problem teaches us that no matter how intimidating an integral looks, a systematic approach—identifying the right substitution, simplifying the trigonometric structure, and applying standard algebraic techniques—will always lead you to the truth.