Hey guys! Are you struggling with Class 12 Maths, especially Chapter 7.2? Don't worry, you're not alone! This chapter, often focusing on integration, can be a bit tricky. But fear not! This guide will walk you through the solutions in Hindi, making it easier to understand and ace your exams.

Understanding Chapter 7.2: Integration

Chapter 7.2 usually delves into various methods of integration. Integration, at its core, is the reverse process of differentiation. Think of it as finding the area under a curve. This chapter typically covers techniques like integration by substitution, integration by parts, and using partial fractions. Each of these techniques is a tool in your arsenal to tackle different types of integrals.

Integration by Substitution: This method is like a clever disguise! You substitute a part of the integrand (the function you're integrating) with a new variable to simplify the integral. It's super useful when you see a function and its derivative hanging around in the integral. For example, if you have an integral with sin(x) and cos(x), you might substitute u = sin(x), then du = cos(x) dx. This transforms the integral into something much easier to handle. Remember, the key is to choose the right substitution – practice makes perfect!

Integration by Parts: This technique is your go-to when you have a product of two functions inside the integral. The formula looks a bit intimidating at first: ∫u dv = uv - ∫v du. But once you get the hang of choosing 'u' and 'dv' correctly, it becomes a powerful tool. The trick is to pick 'u' such that its derivative simplifies the integral. A common strategy is to use the acronym LIATE (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential) to help you prioritize which function to choose as 'u'. The function that comes earlier in the list is usually a good choice for 'u'.

Integration Using Partial Fractions: This method is essential when dealing with rational functions (a fraction where both the numerator and denominator are polynomials). The idea is to break down the complex rational function into simpler fractions that are easier to integrate individually. This involves factoring the denominator and expressing the original fraction as a sum of partial fractions with unknown constants. You then solve for these constants and integrate each partial fraction separately. This technique might seem long, but it's a systematic way to tackle integrals of rational functions.

Understanding these core concepts and practicing different types of problems is crucial for mastering Chapter 7.2.

Key Concepts and Formulas

Before diving into the solutions, let's quickly recap some essential formulas and concepts you'll need:

• Basic Integration Formulas: These are your bread and butter. Make sure you know the integrals of basic functions like x^n, sin(x), cos(x), e^x, and 1/x.
• Integration by Substitution: ∫f'(g(x))g'(x) dx = f(g(x)) + C, where C is the constant of integration.
• Integration by Parts: ∫u dv = uv - ∫v du
• Partial Fractions: Learn how to decompose rational functions into partial fractions based on the factors in the denominator.
• Trigonometric Identities: Knowing your trigonometric identities can greatly simplify integrals involving trigonometric functions.

Keep these formulas handy as you work through the examples. And remember, the constant of integration, 'C', is always important!

Example Problems and Solutions (in Hindi)

Now, let's look at some example problems from Chapter 7.2 and their solutions explained in Hindi. I'll break down each step to make it super clear.

Example 1: Evaluate ∫2x / (1 + x^2) dx

• Solution:
  • Step 1: Notice that the derivative of (1 + x^2) is 2x, which is present in the numerator. This suggests using substitution.
  • Step 2: Let u = 1 + x^2. Then, du = 2x dx.
  • Step 3: Substitute u and du into the integral: ∫(1/u) du.
  • Step 4: Integrate: ln|u| + C.
  • Step 5: Substitute back for u: ln|1 + x^2| + C.
  • Final Answer: ln|1 + x^2| + C

Hindi Explanation:

• Step 1: Yahan dekhiye ki (1 + x^2) ka avkalan 2x hai, jo ansh mein maujood hai. Isse pata chalta hai ki humein substitution ka upayog karna chahiye. (Here, notice that the derivative of (1 + x^2) is 2x, which is present in the numerator. This suggests we should use substitution.)
• Step 2: Maana u = 1 + x^2. Phir, du = 2x dx. (Let u = 1 + x^2. Then, du = 2x dx.)
• Step 3: u aur du ko integral mein substitute karein: ∫(1/u) du. (Substitute u and du into the integral: ∫(1/u) du.)
• Step 4: Integrate karein: ln|u| + C. (Integrate: ln|u| + C.)
• Step 5: u ko phir se substitute karein: ln|1 + x^2| + C. (Substitute back for u: ln|1 + x^2| + C.)
• Final Answer: ln|1 + x^2| + C

Example 2: Evaluate ∫x sin(x) dx

• Solution:
  • Step 1: This is a product of two functions, so we'll use integration by parts.
  • Step 2: Let u = x and dv = sin(x) dx. Then, du = dx and v = -cos(x).
  • Step 3: Apply the integration by parts formula: ∫u dv = uv - ∫v du => ∫x sin(x) dx = -x cos(x) - ∫(-cos(x)) dx
  • Step 4: Simplify and integrate: -x cos(x) + ∫cos(x) dx = -x cos(x) + sin(x) + C
  • Final Answer: -x cos(x) + sin(x) + C

Hindi Explanation:

• Step 1: Yeh do functions ka guna hai, isliye hum integration by parts ka upayog karenge. (This is a product of two functions, so we will use integration by parts.)
• Step 2: Maana u = x aur dv = sin(x) dx. Phir, du = dx aur v = -cos(x). (Let u = x and dv = sin(x) dx. Then, du = dx and v = -cos(x).) Remember LIATE – Algebraic comes before Trigonometric.
• Step 3: Integration by parts formula lagayein: ∫u dv = uv - ∫v du => ∫x sin(x) dx = -x cos(x) - ∫(-cos(x)) dx (Apply the integration by parts formula: ∫u dv = uv - ∫v du => ∫x sin(x) dx = -x cos(x) - ∫(-cos(x)) dx)
• Step 4: Saral karein aur integrate karein: -x cos(x) + ∫cos(x) dx = -x cos(x) + sin(x) + C (Simplify and integrate: -x cos(x) + ∫cos(x) dx = -x cos(x) + sin(x) + C)
• Final Answer: -x cos(x) + sin(x) + C

Example 3: Evaluate ∫1 / (x^2 - 9) dx

• Solution:
  • Step 1: Factor the denominator: x^2 - 9 = (x - 3)(x + 3)
  • Step 2: Use partial fractions: 1 / ((x - 3)(x + 3)) = A / (x - 3) + B / (x + 3)
  • Step 3: Solve for A and B: Multiplying both sides by (x-3)(x+3) gives 1 = A(x+3) + B(x-3). Solving this system yields A = 1/6 and B = -1/6.
  • Step 4: Rewrite the integral: ∫(1 / (x^2 - 9)) dx = ∫(1/6) / (x - 3) dx + ∫(-1/6) / (x + 3) dx
  • Step 5: Integrate: (1/6)ln|x - 3| - (1/6)ln|x + 3| + C
  • Final Answer: (1/6)ln|x - 3| - (1/6)ln|x + 3| + C or (1/6)ln|(x-3)/(x+3)| + C

Hindi Explanation:

• Step 1: Har ko factor karein: x^2 - 9 = (x - 3)(x + 3) (Factor the denominator: x^2 - 9 = (x - 3)(x + 3))
• Step 2: Partial fractions ka upayog karein: 1 / ((x - 3)(x + 3)) = A / (x - 3) + B / (x + 3) (Use partial fractions: 1 / ((x - 3)(x + 3)) = A / (x - 3) + B / (x + 3))
• Step 3: A aur B ke liye solve karein: Dono taraf (x-3)(x+3) se guna karne par 1 = A(x+3) + B(x-3) milta hai. Is system ko solve karne par A = 1/6 aur B = -1/6 milta hai. (Solve for A and B: Multiplying both sides by (x-3)(x+3) gives 1 = A(x+3) + B(x-3). Solving this system yields A = 1/6 and B = -1/6.)
• Step 4: Integral ko phir se likhein: ∫(1 / (x^2 - 9)) dx = ∫(1/6) / (x - 3) dx + ∫(-1/6) / (x + 3) dx (Rewrite the integral: ∫(1 / (x^2 - 9)) dx = ∫(1/6) / (x - 3) dx + ∫(-1/6) / (x + 3) dx)
• Step 5: Integrate karein: (1/6)ln|x - 3| - (1/6)ln|x + 3| + C (Integrate: (1/6)ln|x - 3| - (1/6)ln|x + 3| + C)
• Final Answer: (1/6)ln|x - 3| - (1/6)ln|x + 3| + C or (1/6)ln|(x-3)/(x+3)| + C

Tips for Success

• Practice, Practice, Practice: The more you practice, the better you'll become at recognizing patterns and choosing the right integration techniques. Work through as many problems as possible from your textbook and other resources.
• Understand the Concepts: Don't just memorize formulas! Make sure you understand the underlying concepts behind each integration method. This will help you apply them correctly in different situations.
• Review Basic Differentiation: Since integration is the reverse of differentiation, it's essential to have a strong grasp of differentiation rules.
• Don't Be Afraid to Ask for Help: If you're stuck on a problem, don't hesitate to ask your teacher, classmates, or online resources for help. Sometimes, a fresh perspective can make all the difference.
• Pay Attention to Detail: Integration problems often involve multiple steps, so it's important to be careful and pay attention to detail. A small mistake can lead to a wrong answer.
• Check Your Answers: Whenever possible, check your answers by differentiating the result. If you get back the original integrand, you're on the right track!

Additional Resources

Here are some additional resources that you might find helpful:

• NCERT Textbook: Your NCERT textbook is the primary resource for your Class 12 Maths syllabus. Make sure you understand all the concepts and examples covered in the textbook.
• Online Tutorials: There are many excellent online tutorials and video lectures available on websites like Khan Academy, YouTube, and Vedantu. These resources can provide additional explanations and examples.
• Practice Problems: Look for practice problems online or in supplementary textbooks. The more you practice, the better you'll become at integration.
• Previous Year Question Papers: Solving previous year question papers can help you get a feel for the types of questions that are typically asked in the exams.

Conclusion

So, there you have it! Class 12 Maths Chapter 7.2 solutions in Hindi. Remember, integration might seem daunting at first, but with consistent practice and a solid understanding of the concepts, you can master it. Good luck with your studies, and I hope this guide helps you ace your exams! Keep practicing, and you'll get there. You got this!