Animated Solution for Mathematics - Definite Integration: The area of the region between the curves $y=\sqrt{\frac{1+\sin x}{\cos x}}$ and $y=\sqrt{\frac{1-\sin x}{\cos x}}$ bounded by the lines $x=0$ and $x=\pi /4$ is

Unlocking the Integral

Welcome, fellow traveler on the path to JEE excellence. Today, we are not just solving an integral; we are peeling back the layers of a trigonometric onion.

At first glance, the expression $y=\sqrt{\frac{1+\sin x}{\cos x}}$ looks like a tangled mess of radicals and trigonometric functions. But in the world of advanced calculus, complexity is often just a mask for hidden simplicity.

The Geometric Intuition

Imagine you are standing on the Cartesian plane, looking at the region bounded by two curves between $x=0$ and $x=\pi /4$. We define our area as $A={\int }_0^{\frac{\pi }{4}}(y_1-y_2)dx$.

The challenge here is not the integration itself, but the algebraic form of the integrand. We need to simplify $y_1$ and $y_2$ before we even think about the integral sign.

The Half-Angle Revelation

Let us focus on $y_1=\sqrt{\frac{1+\sin x}{\cos x}}$. The key to unlocking this is the half-angle identity.

We know that $1={\cos }^2(x/2)+{\sin }^2(x/2)$ and $\sin x=2\sin (x/2)\cos (x/2)$. Thus, the numerator becomes $(\cos (x/2)+\sin (x/2){)}^2$.

Similarly, the denominator $\cos x$ is ${\cos }^2(x/2)-{\sin }^2(x/2)$, which is a difference of squares: $(\cos (x/2)-\sin (x/2))(\cos (x/2)+\sin (x/2))$.

When we place these into our radical, the term $(\cos (x/2)+\sin (x/2))$ cancels out, leaving us with:

By dividing both the numerator and denominator by $\cos (x/2)$, we arrive at the elegant form $y_1=\sqrt{\frac{1+\tan (x/2)}{1-\tan (x/2)}}$. By symmetry, $y_2$ becomes $\sqrt{\frac{1-\tan (x/2)}{1+\tan (x/2)}}$.

The Algebraic Symphony

Now, let us define $t=\tan (x/2)$. Our integrand $y_1-y_2$ transforms into $\sqrt{\frac{1+t}{1-t}}-\sqrt{\frac{1-t}{1+t}}$.

When we find a common denominator, the numerator becomes $(1+t)-(1-t)=2t$, and the denominator becomes $\sqrt{(1-t)(1+t)}=\sqrt{1-t^2}$.

Suddenly, the terrifying radical expression has collapsed into the clean, manageable form $\frac{2t}{\sqrt{1-t^2}}$.

The Final Assembly

We are almost there. We have our integrand, but we must also transform our differential $dx$.

Since $t=\tan (x/2)$, we differentiate to find $dt=\frac{1}{2}{\sec }^2(x/2)dx$. Using the identity ${\sec }^2(x/2)=1+t^2$, we find that $dx=\frac{2dt}{1+t^2}$.

Finally, we update our limits. At $x=0$, $t=0$. At $x=\pi /4$, $t=\tan (\pi /8)=\sqrt{2}-1$.

Putting it all together, our area integral becomes:

Look at that result. It is not just an answer; it is a testament to the power of substitution.

You started with a complex trigonometric function and, through careful manipulation, arrived at a structure that reveals the underlying geometry of the problem. Keep this spirit of curiosity alive—every complex problem is just a simple one waiting to be discovered.