Animated Solution for Mathematics - Definite Integration: $I={\int }_{\pi /4}^{\pi /3}\left(\frac{8\sin x-\sin 2x}{x}\right)dx$. Then

Taming the Impossible Integral

Have you ever stared at an integral on a JEE Advanced paper and felt that sinking realization that no standard substitution, no integration by parts, and no trigonometric identity could possibly crack it open? It is a moment of pure panic, but for the seasoned physicist or mathematician, it is actually a moment of opportunity.

Today, we are going to explore a problem that looks like a nightmare but is actually a masterclass in the power of estimation.

The Anatomy of the Integrand

We are presented with the integral:

If you try to integrate this directly, you will find yourself in a loop of frustration. The term $\frac{\sin x}{x}$ is a classic non-elementary integral.

The key, as always, is to look at the numerator. Let us define a function $f(x)=8\sin x-\sin 2x$. Our integral is simply the integral of $\frac{f(x)}{x}$.

The Quadratic Bridge

To understand how $f(x)$ behaves, we need to look at its rate of change. Let us differentiate $f(x)$ with respect to $x$:

Now, we have a mix of $\cos x$ and $\cos 2x$. This is where our trigonometric toolkit comes in. We know the identity $\cos 2x=2{\cos }^{2}x-1$.

Substituting this into our derivative, we get:

This is a beautiful transformation! By substituting $t=\cos x$, we turn a complex trigonometric derivative into a simple quadratic function $g(t)=-4t^2+8t+2$.

The Power of Monotonicity

Now, we must consider our interval for $x$, which is $[\frac{\pi }{4},\frac{\pi }{3}]$. Since $t=\cos x$, and $\cos x$ is a decreasing function, our interval for $t$ becomes $[\cos (\frac{\pi }{3}),\cos (\frac{\pi }{4})]$, which is $[\frac{1}{2},\frac{1}{\sqrt{2}}]$.

Let us analyze the monotonicity of $g(t)$ on this interval. The derivative is $g^{\prime }(t)=-8t+8=8(1-t)$. Since $t$ is at most $\frac{1}{\sqrt{2}}\approx 0.707$, which is less than $1$, $g^{\prime }(t)$ is always positive.

This means $g(t)$ is strictly increasing. The minimum value occurs at the lower bound $t=\frac{1}{2}$, and the maximum occurs at the upper bound $t=\frac{1}{\sqrt{2}}$.

Calculating these values:

So, we have established that $5<f^{\prime }(x)<4\sqrt{2}$.

The Final Bound

We are almost there. By the Mean Value Theorem, since $f(0)=0$, we know that $\frac{f(x)}{x}=f^{\prime }(\xi )$ for some $\xi \in (0,x)$. Because $f^{\prime }(x)$ is increasing, $f^{\prime }(\xi )$ is bounded by the same values as $f^{\prime }(x)$ on our interval.

Thus, we have:

Integrating this inequality across our limits $[\frac{\pi }{4},\frac{\pi }{3}]$ is straightforward:

And there it is! We have successfully bounded the integral without ever needing to find an antiderivative. This is the essence of the JEE Advanced mindset: when the path forward is blocked, look for the geometric or analytical constraints that define the problem. You have mastered the art of estimation.