Animated Solution for Mathematics - Definite Integration: Let $I={\int }_0^1\frac{\sin x}{\sqrt{x}}dx$ and $J={\int }_0^1\frac{\cos x}{\sqrt{x}}dx$. Then which one of the following is true?

Outsmarting the Integral

Welcome, future engineers! Today, we are going to tackle a problem that might look intimidating at first glance. We are faced with two integrals:

If you try to find the antiderivative of $\frac{\sin x}{\sqrt{x}}$, you will quickly realize that you are staring into an abyss. There is no simple function whose derivative is $\frac{\sin x}{\sqrt{x}}$.

But here is the secret of JEE Advanced: when the math gets tough, the smart get clever. We do not need the exact value; we only need to know where these values stand relative to specific numbers. This is the power of estimation.

Taming the Sine Integral

Let us focus on $I={\int }_0^1\frac{\sin x}{\sqrt{x}}dx$. We know that for any positive $x$, the sine function is bounded by the line $y=x$. Specifically, $\sin x<x$.

This is our golden key. Since we are working in the interval $(0,1)$, we know that $\sqrt{x}$ is always positive. This allows us to divide our inequality by $\sqrt{x}$ without any fear of flipping the inequality sign:

Now, imagine the area under the curve $\frac{\sin x}{\sqrt{x}}$ and the area under the curve $\sqrt{x}$. Because the first function is strictly smaller than the second for the entire interval, the area under the first must be strictly smaller than the area under the second.

This is the monotonicity property of integrals. We integrate both sides:

Evaluating the right side is a simple application of the power rule:

Thus, we have proven that $I<\frac{2}{3}$.

Taming the Cosine Integral

Now, let us turn our attention to $J={\int }_0^1\frac{\cos x}{\sqrt{x}}dx$. We need a similar bound. We know that for $x\in (0,1)$, the cosine function is strictly decreasing from $1$.

Therefore, $\cos x<1$. Again, we divide by $\sqrt{x}$ to construct our integrand:

Just as before, we integrate both sides from $0$ to $1$:

The integral of $x^{-1/2}$ is $2\sqrt{x}$. Evaluating this from $0$ to $1$ gives us $2(1)-2(0)=2$. So, we have conclusively shown that $J<2$.

We have successfully navigated the traps. We found that $I<\frac{2}{3}$ and $J<2$.

By using simple inequalities, we have avoided the impossible task of finding an antiderivative and instead arrived at a clear, logical conclusion. This is the essence of physics and mathematics in the JEE: it is not about brute force; it is about finding the most elegant path to the truth.

Keep this mindset, and you will conquer any problem that comes your way!