Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_{-\pi /4}^{\pi /4}\frac{x+\frac{\pi }{4}}{2-\cos 2x}dx$ is

Imagine you are standing before a complex integral, staring at the limits $-\frac{\pi }{4}$ and $\frac{\pi }{4}$. For many students, the first reaction is panic—a desire to dive straight into the algebra.

But as an elite JEE aspirant, you must learn to pause. You must learn to look for the hidden geometry.

The integral we are solving today is:

The moment you see those symmetric limits, your mathematical spider-sense should tingle. This is not just a calculation; it is a test of your ability to recognize symmetry.

The Power of Linearity

We begin by using the property of linearity. We can split this single, intimidating fraction into two distinct integrals: $I=I_1+I_2$, where:

Why do we do this? Because $I_1$ is a gift. Let us define $f(x)=\frac{x}{2-\cos 2x}$.

If we test for symmetry by replacing $x$ with $-x$, we find that $f(-x)=\frac{-x}{2-\cos (-2x)}$. Since cosine is an even function, $\cos (-2x)=\cos 2x$.

Thus, $f(-x)=-f(x)$. This confirms that $f(x)$ is an odd function.

The fundamental theorem of calculus tells us that the integral of an odd function over a symmetric interval $[-a,a]$ is exactly zero. The area on the left of the y-axis perfectly cancels the area on the right. Just like that, $I_1$ vanishes, and our problem is halved.

The Even Transformation

Now we are left with $I_2={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{\frac{\pi }{4}}{2-\cos 2x}dx$. Since the integrand is now an even function, we can use the property ${\int }_{-a}^{a}g(x)dx=2{\int }_0^{a}g(x)dx$.

This simplifies our integral to:

We have successfully stripped away the complexity. Now, we face the core trigonometric challenge: integrating $\frac{1}{2-\cos 2x}$.

The Algebraic Pivot

How do we handle $\cos 2x$ in the denominator? We use the powerful identity $\cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}$.

Substituting this into our integral, we get:

Simplifying the denominator is where precision matters. The expression becomes $2(1+{\tan }^{2}x)-(1-{\tan }^{2}x)=2+2{\tan }^{2}x-1+{\tan }^{2}x=1+3{\tan }^{2}x$.

The term $(1+{\tan }^{2}x)$ in the denominator of the fraction flips to the numerator, becoming ${\sec }^{2}x$. Our integral is now:

The Final Integration

This is the moment of triumph. We see ${\sec }^{2}x$ in the numerator, which is the derivative of $\tan x$.

We let $t=\tan x$, so $dt={\sec }^{2}xdx$. We must also update our limits: when $x=0$, $t=0$; when $x=\frac{\pi }{4}$, $t=1$.

The integral becomes:

This is a standard form: $\int \frac{1}{1+a^2{t}^2}dt=\frac{1}{a}{\tan }^{-1}(at)$. Here, $a=\sqrt{3}$.

Thus, we evaluate:

Plugging in the limits, we get:

The complexity has collapsed into a beautiful, elegant result. Remember, in JEE Advanced, the path is rarely a straight line; it is a series of clever transformations. Keep practicing, and soon, you will see these symmetries before you even pick up your pen.

Final Answer: