CBSE Class 12 Maths: Integrals Exercise 7.7 (NCERT)

Understanding Integrals of Special Irrational Functions

Exercise 7.7 of your NCERT textbook is dedicated to solving integrals that involve a quadratic expression inside a square root. Specifically, we'll be dealing with integrals of the form ∫√(Ax² + Bx + C) dx. The key to solving these is to transform the quadratic expression Ax² + Bx + C into one of three standard forms by using the method of 'completing the square'. This transformation allows us to directly apply specific standard integral formulas that you need to memorise.

These three standard forms are:

1. √(a² - x²): This form arises when the quadratic expression has a positive constant term and a negative x² term, or can be manipulated to this form. For example, ∫√(3 - 2x - x²) dx can be transformed into ∫√(a² - (x+b)²) dx.
2. √(x² + a²): This form appears when both the x² term and the constant term are positive after completing the square, or if the quadratic expression can be reduced to this form. For instance, ∫√(x² + 4x + 5) dx can become ∫√((x+2)² + 1²) dx.
3. √(x² - a²): This form is typically encountered when the x² term is positive and the constant term is negative after completing the square, or if the expression simplifies this way. For example, ∫√(x² - 6x + 7) dx can be rewritten as ∫√((x-3)² - 2²) dx.

The challenge lies in correctly completing the square and then identifying which of the three standard formulas applies. Once identified, direct application of the formula yields the solution, though careful substitution might be needed if the variable is not simply 'x'. Understanding the structure of these formulas and when to apply each is crucial for success in this exercise.

Essential Standard Integral Formulas

Integral of √(a² - x²) dx

The formula for ∫√(a² - x²) dx is (x/2)√(a² - x²) + (a²/2)sin⁻¹(x/a) + C. This is used when the integrand is in the form of a constant squared minus the variable squared, both under a square root.

Integral of √(x² + a²) dx

The formula for ∫√(x² + a²) dx is (x/2)√(x² + a²) + (a²/2)log|x + √(x² + a²)| + C. Apply this when the integrand is in the form of the variable squared plus a constant squared, both under a square root.

Integral of √(x² - a²) dx

The formula for ∫√(x² - a²) dx is (x/2)√(x² - a²) - (a²/2)log|x + √(x² - a²)| + C. This formula is used when the integrand is in the form of the variable squared minus a constant squared, both under a square root. Note the minus sign before the log term.

Worked Examples: Applying the Formulas

• Example 1: Direct Application Evaluate ∫√(9 - x²) dx. Step 1: Identify the form. The integral is of the form ∫√(a² - x²) dx. Here, a² = 9, so a = 3. Step 2: Apply the formula. The formula is (x/2)√(a² - x²) + (a²/2)sin⁻¹(x/a) + C. Step 3: Substitute the value of 'a'. ∫√(9 - x²) dx = (x/2)√(9 - x²) + (9/2)sin⁻¹(x/3) + C. Final Answer: (x/2)√(9 - x²) + (9/2)sin⁻¹(x/3) + C.
• Example 2: Using Completing the Square for √(x² + 4x + 13) dx Evaluate ∫√(x² + 4x + 13) dx. Step 1: Complete the square for the quadratic expression. x² + 4x + 13 = (x² + 4x + 4) - 4 + 13 = (x + 2)² + 9. Step 2: Rewrite the integral with the completed square form. ∫√((x + 2)² + 9) dx. Step 3: Identify the standard form. This is of the form ∫√(y² + a²) dy, where y = x + 2 and a² = 9 (so a = 3). Step 4: Apply the formula for ∫√(y² + a²) dy = (y/2)√(y² + a²) + (a²/2)log|y + √(y² + a²)| + C. Step 5: Substitute back y = x + 2 and a = 3. ∫√((x + 2)² + 9) dx = ((x + 2)/2)√((x + 2)² + 9) + (9/2)log|(x + 2) + √((x + 2)² + 9)| + C. Step 6: Simplify the expression inside the square root. = ((x + 2)/2)√(x² + 4x + 13) + (9/2)log|(x + 2) + √(x² + 4x + 13)| + C. Final Answer: ((x + 2)/2)√(x² + 4x + 13) + (9/2)log|(x + 2) + √(x² + 4x + 13)| + C.
• Example 3: Completing the Square with a negative x² term Evaluate ∫√(5 - 4x - x²) dx. Step 1: Complete the square. Factor out -1 from the quadratic terms inside the square root, then complete the square. 5 - 4x - x² = 5 - (x² + 4x) = 5 - (x² + 4x + 4 - 4) = 5 - ((x + 2)² - 4) = 5 - (x + 2)² + 4 = 9 - (x + 2)². Step 2: Rewrite the integral. ∫√(9 - (x + 2)²) dx. Step 3: Identify the standard form. This is of the form ∫√(a² - y²) dy, where a² = 9 (a = 3) and y = x + 2. Step 4: Apply the formula for ∫√(a² - y²) dy = (y/2)√(a² - y²) + (a²/2)sin⁻¹(y/a) + C. Step 5: Substitute back y = x + 2 and a = 3. ∫√(9 - (x + 2)²) dx = ((x + 2)/2)√(9 - (x + 2)²) + (9/2)sin⁻¹((x + 2)/3) + C. Step 6: Simplify the expression inside the square root. = ((x + 2)/2)√(5 - 4x - x²) + (9/2)sin⁻¹((x + 2)/3) + C. Final Answer: ((x + 2)/2)√(5 - 4x - x²) + (9/2)sin⁻¹((x + 2)/3) + C.

Exam Tips and Common Mistakes

Mastering integrals of special irrational functions requires attention to detail. Here are some crucial tips and common pitfalls to avoid:

• Perfecting Completing the Square: This is the most critical first step. A small error here will lead to an incorrect final answer. Remember, for a quadratic ax² + bx + c, the process is a(x² + (b/a)x) + c. If a is not 1, factor it out first. If it's c - ax² - bx, factor out -a from ax² + bx before completing the square.
• Formula Recall: There's no shortcut here; you MUST memorize the three standard integral formulas. Pay close attention to the + or - signs, especially in the log term for √(x² - a²) dx and the sin⁻¹ term for √(a² - x²) dx.
• Identifying 'a' and 'x' (or 'y'): After completing the square, clearly identify what a represents (a constant) and what x (or the transformed variable y = x + k) represents in your standard formula. Sometimes, the variable might be 2x or 3x, which requires a substitution (u = 2x or u = 3x) before applying the standard formula.
• Simplification: Always simplify the expression y² ± a² back to the original quadratic Ax² + Bx + C in your final answer's square root term. This makes the answer cleaner and easier to verify.

Practice Questions with Solutions

• Q: Evaluate ∫√(x² + 6x + 1) dx. A: Step 1: Complete the square for x² + 6x + 1. x² + 6x + 1 = (x² + 6x + 9) - 9 + 1 = (x + 3)² - 8. Step 2: Rewrite the integral as ∫√((x + 3)² - 8) dx. Step 3: Identify the form ∫√(y² - a²) dy, where y = x + 3 and a² = 8 (a = √8 = 2√2). Step 4: Apply the formula (y/2)√(y² - a²) - (a²/2)log|y + √(y² - a²)| + C. Step 5: Substitute y = x + 3 and a = 2√2. = ((x + 3)/2)√((x + 3)² - 8) - (8/2)log|(x + 3) + √((x + 3)² - 8)| + C. Step 6: Simplify. = ((x + 3)/2)√(x² + 6x + 1) - 4log|(x + 3) + √(x² + 6x + 1)| + C. Final answer: ((x + 3)/2)√(x² + 6x + 1) - 4log|(x + 3) + √(x² + 6x + 1)| + C.
• Q: Evaluate ∫√(1 - 2x - x²) dx. A: Step 1: Complete the square for 1 - 2x - x². 1 - 2x - x² = 1 - (x² + 2x) = 1 - (x² + 2x + 1 - 1) = 1 - ((x + 1)² - 1) = 1 - (x + 1)² + 1 = 2 - (x + 1)². Step 2: Rewrite the integral as ∫√(2 - (x + 1)²) dx. Step 3: Identify the form ∫√(a² - y²) dy, where a² = 2 (a = √2) and y = x + 1. Step 4: Apply the formula (y/2)√(a² - y²) + (a²/2)sin⁻¹(y/a) + C. Step 5: Substitute y = x + 1 and a = √2. = ((x + 1)/2)√(2 - (x + 1)²) + (2/2)sin⁻¹((x + 1)/√2) + C. Step 6: Simplify. = ((x + 1)/2)√(1 - 2x - x²) + sin⁻¹((x + 1)/√2) + C. Final answer: ((x + 1)/2)√(1 - 2x - x²) + sin⁻¹((x + 1)/√2) + C.
• Q: Evaluate ∫√(4x² + 9) dx. A: Step 1: Factor out 4 from 4x² + 9 inside the square root. ∫√(4(x² + 9/4)) dx = ∫2√(x² + (3/2)²) dx. Step 2: Identify the form ∫√(x² + a²) dx, where a = 3/2. Step 3: Apply the formula 2 [(x/2)√(x² + a²) + (a²/2)log|x + √(x² + a²)|] + C. Step 4: Substitute a = 3/2. = 2 [(x/2)√(x² + (3/2)²) + ((3/2)²/2)log|x + √(x² + (3/2)²)|] + C. = 2 [(x/2)√(x² + 9/4) + (9/8)log|x + √(x² + 9/4)|] + C. Step 5: Distribute the 2 and simplify. = x√(x² + 9/4) + (9/4)log|x + √(x² + 9/4)| + C. Final answer: x√(x² + 9/4) + (9/4)log|x + √(x² + 9/4)| + C. (Alternatively, x/2 √(4x^2+9) + 9/4 log |x + 1/2 √(4x^2+9)| + C)
• Q: Evaluate ∫√(x² - 10x + 21) dx. A: Step 1: Complete the square for x² - 10x + 21. x² - 10x + 21 = (x² - 10x + 25) - 25 + 21 = (x - 5)² - 4. Step 2: Rewrite the integral as ∫√((x - 5)² - 4) dx. Step 3: Identify the form ∫√(y² - a²) dy, where y = x - 5 and a² = 4 (a = 2). Step 4: Apply the formula (y/2)√(y² - a²) - (a²/2)log|y + √(y² - a²)| + C. Step 5: Substitute y = x - 5 and a = 2. = ((x - 5)/2)√((x - 5)² - 4) - (4/2)log|(x - 5) + √((x - 5)² - 4)| + C. Step 6: Simplify. = ((x - 5)/2)√(x² - 10x + 21) - 2log|(x - 5) + √(x² - 10x + 21)| + C. Final answer: ((x - 5)/2)√(x² - 10x + 21) - 2log|(x - 5) + √(x² - 10x + 21)| + C.

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