Exercise 9.3 - Definite Integrals
(i) ∫04 dx/(x² - 4)
1. Factor the denominator: x² - 4 = (x - 2)(x + 2)
2. Perform partial fraction decomposition:
1/(x² - 4) = A/(x - 2) + B/(x + 2)
3. Solve for A and B: A = 1/4, B = -1/4
4. Rewrite the integral:
∫ (1/4)/(x - 2) dx - ∫ (1/4)/(x + 2) dx
5. Integrate:
(1/4)ln|x - 2| - (1/4)ln|x + 2| + C
6. Evaluate from 0 to 4:
= (1/4)[ln2 - ln6] - (1/4)[ln2 - ln2]
= (1/4)ln(1/3) ≈ -0.2747
(ii) ∫-11 dx/(x² + 2x + 5)
1. Complete the square in denominator:
x² + 2x + 5 = (x + 1)² + 4
2. Recognize standard integral form:
∫ du/(u² + a²) = (1/a)tan⁻¹(u/a) + C
3. Let u = x + 1, du = dx, a = 2
4. Rewrite integral:
∫ du/(u² + 4)
5. Integrate:
(1/2)tan⁻¹(u/2) + C
6. Evaluate from u=0 to u=2:
= (1/2)[tan⁻¹(1) - tan⁻¹(0)] = (1/2)(π/4 - 0) = π/8
(iii) ∫01 √[(1 - x)/(1 + x)] dx
1. Rationalize the integrand:
√[(1 - x)²/(1 - x²)] = (1 - x)/√(1 - x²)
2. Split into two integrals:
∫ 1/√(1 - x²) dx - ∫ x/√(1 - x²) dx
3. First integral is standard:
∫ 1/√(1 - x²) dx = sin⁻¹x + C
4. Second integral by substitution (u = 1 - x²):
∫ x/√(1 - x²) dx = -√(1 - x²) + C
5. Combine results:
sin⁻¹x + √(1 - x²) + C
6. Evaluate from 0 to 1:
= (π/2 + 0) - (0 + 1) = π/2 - 1
(iv) ∫0π/2 eˣ(1 + sinx)/(1 + cosx) dx
1. Split the integrand:
eˣ(1/(1 + cosx) + sinx/(1 + cosx))
2. Simplify using trigonometric identities:
1/(1 + cosx) = (1 - cosx)/sin²x
sinx/(1 + cosx) = tan(x/2)
3. Rewrite integral:
∫ eˣ[(1 - cosx)/sin²x + tan(x/2)] dx
4. Recognize integration by parts pattern:
∫ eˣ[f(x) + f'(x)] dx = eˣf(x) + C
5. Here f(x) = tan(x/2), f'(x) = (1/2)sec²(x/2)
6. Final answer:
eˣtan(x/2) evaluated from 0 to π/2 = eπ/2(1) - e0(0) = eπ/2
(v) ∫0π/2 √[cosθ sin³θ] dθ
1. Rewrite integrand:
√[cosθ sin³θ] = cos1/2θ sin3/2θ
2. Use trigonometric identity:
= sinθ (cosθ)1/2(sinθ)1/2
3. Let u = sinθ, du = cosθ dθ
4. Transform integral:
∫ u3/2(1 - u)-1/2 du
5. Recognize Beta function form
6. Final answer:
Γ(5/2)Γ(1/2)/Γ(3) = (3√π/4)(√π)/(2) = 3π/8
(vi) ∫01 (1 - x²)/(1 + x²)² dx
1. Break into two integrals:
∫ 1/(1 + x²)² dx - ∫ x²/(1 + x²)² dx
2. First integral by trigonometric substitution (x = tanθ)
3. Second integral by parts:
u = x, dv = x/(1 + x²)² dx
4. After calculation:
First integral = [x/(2(1 + x²)) + (1/2)tan⁻¹x]
5. Second integral = [-x/(2(1 + x²)) + (1/2)tan⁻¹x]
6. Combine and evaluate:
Final answer = π/4 - 1/2
Properties of Integration
(i) ∫-55 x cos[(eˣ - 1)/(eˣ + 1)] dx
1. Check if function is odd:
f(-x) = -x cos[(e⁻ˣ - 1)/(e⁻ˣ + 1)] = -x cos[(1 - eˣ)/(1 + eˣ)] = -x cos[-(eˣ - 1)/(eˣ + 1)] = -x cos[(eˣ - 1)/(eˣ + 1)] = -f(x)
2. Therefore, the integrand is odd
3. Integral of odd function over symmetric limits is zero
(ii) ∫π/2π/2 (x⁵ + x cosx + tan³x + 1) dx
1. Notice the limits are identical (π/2 to π/2)
2. Any definite integral with identical limits is zero
(iii) ∫-π/4π/4 sin²x dx
1. Use identity: sin²x = (1 - cos2x)/2
2. Rewrite integral:
(1/2)∫ (1 - cos2x) dx
3. Integrate:
(1/2)[x - (1/2)sin2x]
4. Evaluate from -π/4 to π/4:
= (1/2)[(π/4 - 1/2) - (-π/4 + 1/2)] = (1/2)(π/2 - 1) = π/4 - 1/2
(iv) ∫02π x log[(3 + cosx)/(3 - cosx)] dx
1. Let I = ∫02π x f(cosx) dx
2. Use property: ∫02π x f(cosx) dx = π ∫02π f(cosx) dx when f is even
3. Here f(cosx) = log[(3 + cosx)/(3 - cosx)]
4. The function is odd since f(-t) = -f(t)
5. Therefore, the integral evaluates to zero
(v) ∫02π sin⁴x cos³x dx
1. Observe that cos³x is odd and sin⁴x is even
2. The product is odd since even × odd = odd
3. Over symmetric interval [0, 2π], integral of odd function is zero
(vi) ∫01 |5x - 3| dx
1. Find where 5x - 3 changes sign:
5x - 3 = 0 ⇒ x = 3/5 = 0.6
2. Split integral at x = 0.6
3. For x ∈ [0, 0.6], |5x - 3| = 3 - 5x
4. For x ∈ [0.6, 1], |5x - 3| = 5x - 3
5. Compute two integrals:
∫00.6 (3 - 5x) dx + ∫0.61 (5x - 3) dx
6. First integral:
[3x - (5/2)x²]00.6 = 1.8 - 0.9 = 0.9
7. Second integral:
[(5/2)x² - 3x]0.61 = (2.5 - 3) - (0.9 - 1.8) = 0.4
8. Final answer:
0.9 + 0.4 = 1.3