Sigma Percentile
JEE Advanced 2005
LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: Evaluate .

Visualized Solution

Analyzing the Integral

  • Original Integral:
  • Notice the repeating terms.
  • The presence of hints at a substitution.

Substitution:

  • Let
  • Differentiating both sides with respect to :

Updating Integration Limits

  • Lower limit: When
  • Upper limit: When
  • The integral becomes:

Flipping Limits of Integration

  • Using the property:
  • The negative sign from flips the limits.

Splitting into Two Integrals

  • Let
  • Symmetric limits suggest checking for odd/even functions.

Evaluating the Odd Part ()

  • Let
  • Since is an odd function,

Evaluating the Even Part ()

  • Let
  • Since is even,
  • For , , so

Standard Integral Formula

  • We need to evaluate:
  • Standard Formula:
  • In our case: and

Substituting and

  • Antiderivative:

Evaluating the Definite Integral

  • Upper limit ():
  • Lower limit ():

Final Answer

  • Combining both limits:
  • Factoring out :

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we stand before an integral that, at first glance, seems designed to intimidate:
It contains exponentials, absolute values, and nested trigonometric functions. But remember, in the world of JEE Advanced, complexity is often just a mask for elegance. Our job is to peel back that mask.

The Power of Substitution

Whenever you see a function and its derivative sitting together, your intuition should immediately fire. Look at the term at the end of our integral; it is the derivative of , which appears repeatedly inside the other functions.
Let . Then, , or . This substitution is the key that unlocks the door.
When , . When , . Our integral transforms into:

The Beauty of Symmetry

Now, let us clean up. That negative sign from is a gift; we can use it to flip the limits of integration from to .
Now we have:
Whenever you see limits from to , your brain should scream: "Check for even and odd functions!" This is a favorite JEE tactic.
Let us split the integral into two parts:
For , the function is odd because . The integral of an odd function over symmetric limits is zero, as the positive and negative areas cancel out perfectly. We are left only with .

The Final Integration

Since is an even function, we can simplify to:
We are now left with a standard integral. We use the standard formula:
Here, and . The denominator . The integral becomes:
Evaluating at the limits, we get:
This simplifies to our final, elegant answer:
See? The complexity was just a test of your patience and your ability to spot the underlying symmetry. Keep practicing, and these patterns will become second nature!

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