Animated Solution for Mathematics - Indefinite Integration: If $f^{\prime }(x)={\tan }^{-1}(\sec x+\tan x),-\frac{\pi }{2}<x<\frac{\pi }{2}$, and $f(0)=0$, then $f(1)$ is equal to:

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of inverse trigonometry.

We are given $f^{\prime }(x)={\tan }^{-1}(\sec x+\tan x)$ and asked to find $f(1)$ given $f(0)=0$. Many students see this and immediately panic, thinking they need to perform a grueling integration of an inverse tangent function.

But here is the secret of JEE Advanced: the problem is not about integration; it is about simplification. Let us embark on this journey together.

Our first objective is to tame the beast inside the inverse tangent function: $\sec x+\tan x$. This expression is a classic disguise.

In the world of trigonometry, when you see secants and tangents, your first instinct should be to strip them down to their fundamental components: sine and cosine. We know that $\sec x=\frac{1}{\cos x}$ and $\tan x=\frac{\sin x}{\cos x}$.

When we combine these, we get:

This is much cleaner, but we are not done yet. We have a numerator of $1+\sin x$ and a denominator of $\cos x$. This structure is screaming for half-angle identities.

Now, let us perform some surgical precision. We want to turn $1+\sin x$ into a perfect square.

Recall the fundamental identity $1={\cos }^2(x/2)+{\sin }^2(x/2)$ and the double-angle identity $\sin x=2\sin (x/2)\cos (x/2)$. If we substitute these into our numerator, we get:

This is the expansion of $(a+b{)}^2$. So, our numerator becomes $(\cos (x/2)+\sin (x/2){)}^2$.

Now, look at the denominator, $\cos x$. Using the double-angle identity for cosine, we have:

This is a difference of squares, which factors beautifully into $(\cos (x/2)-\sin (x/2))(\cos (x/2)+\sin (x/2))$.

When we put the numerator and denominator back together, something magical happens:

Notice the common factor? We can cancel one $(\cos (x/2)+\sin (x/2))$ from the top and bottom. We are left with:

To make this look like a tangent function, we divide both the numerator and denominator by $\cos (x/2)$. This gives us:

This is the standard expansion for $\tan (\frac{\pi }{4}+\frac{x}{2})$. Our derivative has transformed from a terrifying inverse tangent expression into:

Before we celebrate, we must be careful. We are given the domain $x\in (-\frac{\pi }{2},\frac{\pi }{2})$.

We need to ensure that the angle $\frac{\pi }{4}+\frac{x}{2}$ is within the principal domain of the inverse tangent function, which is $(-\frac{\pi }{2},\frac{\pi }{2})$. If $x$ is between $-\frac{\pi }{2}$ and $\frac{\pi }{2}$, then $x/2$ is between $-\frac{\pi }{4}$ and $\frac{\pi }{4}$.

Adding $\frac{\pi }{4}$ to this range gives us an interval of $(0,\frac{\pi }{2})$. This is perfectly within the principal domain! Therefore, we can safely say:

Now, the hard part is over. We have a simple linear derivative: $f^{\prime }(x)=\frac{\pi }{4}+\frac{x}{2}$.

To find $f(x)$, we integrate with respect to $x$:

We are given the initial condition $f(0)=0$. Plugging in $x=0$, we find $0=0+0+C$, so $C=0$.

Our final function is $f(x)=\frac{\pi x}{4}+\frac{x^2}{4}$. To find $f(1)$, we simply substitute $x=1$:

And there you have it! We started with a complex trigonometric expression and, through careful manipulation and identity usage, arrived at a simple, elegant solution. This is the essence of JEE Advanced mathematics: patience, precision, and the courage to simplify.