Animated Solution for Mathematics - Indefinite Integration: Let $I(x)=\int \frac{x^2(x{\sec }^{2}x+\tan x)}{(x\tan x+1{)}^2}dx$. If $I(0)=0$, then $I\left(\frac{\pi }{4}\right)$ is equal to

Welcome, fellow explorer of the mathematical universe! Today, we are standing before an integral that, at first glance, looks like a chaotic mess of trigonometric functions and algebraic terms.

We have the following integral:

It is easy to feel intimidated by such an expression, but remember: in the world of JEE Advanced, complexity is often just a mask for elegance. Our mission is to peel back that mask.

Finding the Substitution

Whenever you encounter a complex fraction in an integral, your first instinct should be to investigate the denominator. Let us look at the expression inside the square: $u=x\tan x+1$.

What happens if we differentiate this with respect to $x$? Using the product rule on $x\tan x$, we get:

The derivative of the constant $1$ is simply $0$. Therefore, $\frac{du}{dx}=x{\sec }^{2}x+\tan x$.

Look at that! The derivative is exactly the term sitting in our numerator. This is not a coincidence; it is the key to the entire problem.

Integration by Parts

With this discovery, we can employ the powerful tool of Integration by Parts. We need to split our integrand into two functions.

We choose $f(x)=x^2$ because it is a polynomial that simplifies beautifully when differentiated. The remaining part, $g(x)=\frac{x{\sec }^{2}x+\tan x}{(x\tan x+1{)}^2}$, becomes our second function.

We already know how to integrate $g(x)$ because we just found that its numerator is the derivative of the base of its denominator. The integral of $\frac{du}{u^2}$ is $-\frac{1}{u}$.

Thus, we have:

Applying the Integration by Parts formula, $\int f\cdot gdx=f\int gdx-\int (f^{\prime }\int gdx)dx$, we get:

The two negative signs cancel out, leaving us with:

Now, we are left with a new, simpler integral. Let us call it $I_1=\int \frac{x}{x\tan x+1}dx$.

By converting $\tan x$ to $\frac{\sin x}{\cos x}$ and multiplying the numerator and denominator by $\cos x$, we get:

If we set $t=x\sin x+\cos x$, then $dt=(x\cos x+\sin x-\sin x)dx=x\cos xdx$. The numerator is once again the derivative of the denominator!

This gives us $I_1=\ln |x\sin x+\cos x|$.

Putting it all together, our indefinite integral is:

We are given $I(0)=0$. Plugging in $x=0$, we find that $C=0$.

Finally, we evaluate at $x=\frac{\pi }{4}$. Using the values $\tan \frac{\pi }{4}=1$, $\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}$, and $\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$, the expression simplifies to:

You have successfully navigated the complexity and arrived at the solution. Keep practicing, and remember that every difficult integral is just a puzzle waiting for your insight!