Animated Solution for Mathematics - Definite Integration: Match the integrals in Column I with the values in Column II.

Welcome, future engineers! Today, we are going to dissect a classic JEE Advanced matching problem. These problems are not just about calculation; they are about pattern recognition and the elegance of standard forms.

When you look at an integral, you shouldn't immediately reach for complex substitutions. Instead, train your eyes to see the underlying structure. Let's embark on this journey through the four integrals provided.

The Familiar Friends

Let's start with Integral (A):

This is the quintessential inverse trigonometric form. We know that the derivative of ${\tan }^{-1}(x)$ is $\frac{1}{1+x^2}$.

Therefore, the integral is simply ${\tan }^{-1}(x)$. When we apply the limits from $-1$ to $1$, we get ${\tan }^{-1}(1)-{\tan }^{-1}(-1)$.

Since ${\tan }^{-1}(1)=\frac{\pi }{4}$ and ${\tan }^{-1}(-1)=-\frac{\pi }{4}$, the result is:

Next, consider Integral (B):

This is another standard form, the derivative of ${\sin }^{-1}(x)$. Applying the limits from $0$ to $1$, we get ${\sin }^{-1}(1)-{\sin }^{-1}(0)=\frac{\pi }{2}-0=\frac{\pi }{2}$.

It is fascinating to see that both (A) and (B) map to the same value, $\frac{\pi }{2}$. This is a gentle reminder that in matching problems, multiple paths can lead to the same destination.

The Logarithmic Twist

Now, let's tackle Integral (C):

Here is where many students stumble. Do not confuse this with ${\tan }^{-1}(x)$; the presence of the minus sign changes everything.

We must use the standard formula:

With $a=1$, our integral becomes:

Substituting the upper limit $x=3$, we get $\log |\frac{1+3}{1-3}|=\log |\frac{4}{-2}|=\log (2)$. Substituting the lower limit $x=2$, we get $\log |\frac{1+2}{1-2}|=\log |\frac{3}{-1}|=\log (3)$.

The final result is $\frac{1}{2}[\log (2)-\log (3)]$. Using the property $\log (m)-\log (n)=\log (\frac{m}{n})$, we arrive at:

The Secant Inverse

Finally, we look at Integral (D):

This is the standard form for ${\sec }^{-1}(x)$. When we evaluate this from $1$ to $2$, we get ${\sec }^{-1}(2)-{\sec }^{-1}(1)$.

We know that ${\sec }^{-1}(2)=\frac{\pi }{3}$ and ${\sec }^{-1}(1)=0$. Thus, the result is:

Mastery Through Practice

We have successfully navigated through these four integrals. By recognizing the standard forms—${\tan }^{-1}(x)$, ${\sin }^{-1}(x)$, the logarithmic form, and ${\sec }^{-1}(x)$—we turned what could have been a tedious calculation into a series of elegant steps.

Remember, the key to JEE success is not just knowing the formulas, but knowing when to apply them. Keep practicing, stay curious, and keep pushing your boundaries!