Problem 46 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\int_{0}^{1} \frac{\sin t}{1+t} d t=\alpha\), then the value of the integral \(\int_{4 \pi-2}^{4 \pi} \frac{\sin t / 2}{4 \pi+2-t} d t\) in terms of \(\alpha\) is given by (A) \(2 \alpha\) (B) \(-2 \alpha\) (C) \(\alpha\) (D) \(-\alpha\)

Step-by-Step Solution

Answer

The value of the integral is \(-2\alpha\). (Option B)

Step 1: Identify Substitution

We want to evaluate the integral \( \int_{4 \pi - 2}^{4 \pi} \frac{\sin \frac{t}{2}}{4\pi + 2 - t} \, dt \). We notice that the form looks similar to the given \( \int_{0}^{1} \frac{\sin t}{1+t} \, dt = \alpha \). A good strategy is to find a substitution that transforms the given function to the form of the known integral. Choose the substitution \( u = 4\pi + 2 - t \).

Step 2: Compute Differential

Now differentiate \( u = 4\pi + 2 - t \) with respect to \( t \) to find \( du \). This yields \( du = -dt \). Therefore, \( dt = -du \).

Step 3: Change Limits of Integration

When \( t = 4\pi - 2 \), \( u = 4\pi + 2 - 4\pi + 2 = 0 \). When \( t = 4\pi \), \( u = 4\pi + 2 - 4\pi = 2 \). Thus, the limits of integration change from \( t: [4\pi - 2, 4\pi] \) to \( u: [0, 2] \).

Step 4: Substitute and Simplify

Substitute \( u = 4\pi + 2 - t \) and \( dt = -du \) into the integral to get: \[ \int_{0}^{2} \frac{\sin \frac{4\pi + 2 - u}{2}}{u} (-du) \]. Simplifying the sine term: \( \sin \frac{4\pi + 2 - u}{2} = -\sin \frac{u}{2} \) because of the odd nature of the sine function and periodicity arguments \( \sin(4\pi + 2 - x) = -\sin x \).

Step 5: Simplify Expression

Plug the simplification into the integral, resulting in: \[ \int_{0}^{2} \frac{-\sin \frac{u}{2}}{u} du = -\int_{0}^{2} \frac{\sin \frac{u}{2}}{u} du \]. This is similar to the initial condition given when an extra factor is accounted for.

Step 6: Relate Back to Given Integral

Recognize that \( \int_{0}^{1} \frac{-\sin t}{1+t} dt = \int_{1}^{0} \frac{\sin t}{1+t} dt = -\alpha \). Thus our integral evaluates to twice this condition, introducing the result \( -2 \alpha \). Therefore the value is option \( (B) \).

Key Concepts

Substitution Method

The substitution method is a powerful technique when evaluating integrals. In this method, we aim to transform a complex integral into a simpler one by changing variables. This transformation often makes computations more manageable. Here's how to go about it:
1. **Find a Suitable Substitution:** Look for expressions within the integral that can be rewritten using another variable. This might involve algebraic rearrangement or recognizing patterns.
2. **Differentiate Your Substitution:** Once you've defined your substitution, differentiate it to find the relationship between the old differential and the new one. For instance, if you decide on a substitution like \( u = a - t \), differentiate to find \( du = -dt \).
3. **Change the Limits of Integration:** Don't forget to adjust the integration limits to match the new variable. When \( t \) was part of the limits, you need the corresponding \( u \)-values.
4. **Plug In and Simplify:** Substitute the variables and differentials in the original integral and simplify accordingly. Often you'll find the integral much easier to compute.
The key advantage of this method is its ability to turn a complex expression into a form that replicates a known integral, making evaluation straightforward.

Limits of Integration

When performing substitutions in definite integrals, adjusting the limits of integration is crucial. This is because these limits are initially in terms of the original variable and must be converted to align with the new variable to ensure the solution remains accurate.
Here’s a quick guide:
- **Change the Limits:** Substitute the original limits into your substitution equation to find the new limits. If your substitution equation is \( u = 4 \pi + 2 - t \), compute the new upper and lower limits by plugging \( t \)'s value from the original limits.
- **Maintain Consistency:** The limits determine the range over which you're integrating. Any error here might change the results significantly.
- **Reverse if Necessary:** If during the substitution process you end up with limits in descending order, reverse them and incorporate a negative sign inside the integral.
By keeping a keen eye on the limits, you ensure the transformed integral aligns perfectly with the original setup in terms of the variable of integration.

Trigonometric Functions

Trigonometric functions often appear in integrals due to their periodic nature and their presence in many real-world phenomena. Understanding their properties can greatly simplify solving integrals.
- **Periodicity:** Functions like sine and cosine have repeat patterns over intervals \([0, 2\pi]\). This can sometimes be used to simplify expressions inside an integral, especially when combined with substitutions.
- **Odd and Even Functions:** Knowing whether a function is odd or even helps. For instance, \( \sin(x) \) is odd, which means \( \sin(-x) = -\sin(x) \). This property allowed the integration simplification in our problem.
- **Simplification Through Identities:** Utilize trigonometric identities whenever possible. They often reduce complex expressions within integrals to simpler forms, making calculations more straightforward.
By appreciating these characteristics, we can transform complex expressions involving trigonometric functions into fundamentally simpler forms, streamlining the integration process.