Exercise 7.9 answers: Integrals

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1. Let t = x² + 1, so dt = 2x dx, i.e. x dx = dt/2.
2. When x = 0, t = 1; when x = 1, t = 2.
3. The integral becomes (1/2) ∫₁² dt/t.
4. This is (1/2)[log t] from 1 to 2 = (1/2)(log 2 − log 1).
5. Since log 1 = 0, the value is (1/2) log 2.

Answer(log 2)/2

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1. Let t = sin φ, so dt = cos φ dφ.
2. Write cos⁵ φ dφ as cos⁴ φ (cos φ dφ) = (1 − sin² φ)² dt = (1 − t²)² dt.
3. When φ = 0, t = 0; when φ = π/2, t = 1.
4. The integral becomes ∫₀¹ √t (1 − t²)² dt = ∫₀¹ (t^(1/2) − 2t^(5/2) + t^(9/2)) dt.
5. Integrate term by term: (2/3)t^(3/2) − (4/7)t^(7/2) + (2/11)t^(11/2), from 0 to 1.
6. At t = 1: 2/3 − 4/7 + 2/11 = 64/231 (common denominator 231).
7. At t = 0, every term is 0, so the value is 64/231.

Answer64/231

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1. Put x = tan θ. Then 2x/(1 + x²) = sin 2θ, so sin⁻¹(2x/(1+x²)) = 2θ for x between 0 and 1.
2. When x = 0, θ = 0; when x = 1, θ = π/4.
3. Also dx = sec² θ dθ, so the integral becomes ∫₀^(π/4) 2θ sec² θ dθ.
4. Integrate by parts with u = 2θ, dv = sec² θ dθ, v = tan θ: this gives [2θ tan θ] − ∫ 2 tan θ dθ.
5. [2θ tan θ] from 0 to π/4 is 2(π/4)(1) − 0 = π/2.
6. ∫₀^(π/4) 2 tan θ dθ = [−2 log(cos θ)] from 0 to π/4 = −2 log(1/√2) = log 2.
7. So the value is π/2 − log 2.

Answerπ/2 − log 2

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1. Evaluate: x√(x + 2) dx (Put x + 2 = t²).
2. Let x + 2 = t², so x = t² − 2 and dx = 2t dt.
3. When x = 0, t = √2; when x = 2, t = 2.
4. x√(x+2) dx becomes (t² − 2)(t)(2t dt) = 2(t⁴ − 2t²) dt.
5. Integrate: 2[t⁵/5 − 2t³/3].
6. At t = 2: 2[32/5 − 16/3] = 32/15.
7. At t = √2: 2[4√2/5 − 4√2/3] = −16√2/15.
8. Subtract: value = 32/15 − (−16√2/15) = 32/15 + 16√2/15.

Answer32/15 + (16√2)/15

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1. Let t = cos x, so dt = −sin x dx.
2. When x = 0, t = 1; when x = π/2, t = 0 — the new limits come out the other way round.
3. The integral becomes −∫₁⁰ dt/(1+t²) = ∫₀¹ dt/(1+t²).
4. This is [tan⁻¹ t] from 0 to 1 = tan⁻¹1 − tan⁻¹0 = π/4 − 0.
5. Value = π/4.

Answerπ/4

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1. Rearrange the denominator: x + 4 − x² = 4 + x − x² = −(x² − x − 4).
2. Complete the square: x² − x − 4 = (x − 1/2)² − 1/4 − 4 = (x − 1/2)² − 17/4. So 4 + x − x² = (√17/2)² − (x − 1/2)².
3. Use ∫ du/(a² − u²) = (1/(2a)) log|(a + u)/(a − u)| with a = √17/2 and u = x − 1/2. Here 1/(2a) = 1/√17.
4. Antiderivative: (1/√17) log[(√17 − 1 + 2x)/(√17 + 1 − 2x)]. Both brackets stay positive for x from 0 to 2, so the bars can be dropped.
5. At x = 2: (1/√17) log[(√17 + 3)/(√17 − 3)]. At x = 0: (1/√17) log[(√17 − 1)/(√17 + 1)].
6. Subtract, using log p − log q = log(p/q): value = (1/√17) log[(√17 + 3)(√17 + 1) / ((√17 − 3)(√17 − 1))].
7. Multiply out: (√17 + 3)(√17 + 1) = 20 + 4√17 and (√17 − 3)(√17 − 1) = 20 − 4√17. Dividing top and bottom by 4 leaves (5 + √17)/(5 − √17).
8. So the value is (1/√17) log[(5 + √17)/(5 − √17)], which can also be written (1/√17) log[(21 + 5√17)/4].

Answer(1/√17) log[(5 + √17)/(5 − √17)]

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1. Complete the square: x² + 2x + 5 = (x+1)² + 4.
2. Use ∫ dx/(u²+a²) = (1/a) tan⁻¹(u/a), with a = 2, u = x+1.
3. Antiderivative: (1/2) tan⁻¹((x+1)/2).
4. At x = 1: (1/2) tan⁻¹(1) = π/8; at x = −1: (1/2) tan⁻¹(0) = 0.
5. Value = π/8.

Answerπ/8

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1. Let t = 2x, so dx = dt/2. When x = 1, t = 2; when x = 2, t = 4.
2. Then 1/x = 2/t and 1/(2x²) = 2/t², so the integrand becomes (2/t − 2/t²) eᵗ · (dt/2) = eᵗ (1/t − 1/t²) dt.
3. Here f(t) = 1/t and f′(t) = −1/t², so the integrand is eᵗ[f(t) + f′(t)].
4. Use ∫ eᵗ[f(t) + f′(t)] dt = eᵗ f(t): the antiderivative is eᵗ/t.
5. Evaluate from t = 2 to t = 4: e⁴/4 − e²/2.

Answere⁴/4 − e²/2

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1. Write (x − x³)^(1/3) = x^(1/3)(1 − x²)^(1/3), so the integrand becomes (1 − x²)^(1/3)/x^(11/3).
2. Factor x² out of (1 − x²): 1 − x² = x²(1/x² − 1), so the integrand simplifies to (1/x² − 1)^(1/3)/x³.
3. Let t = 1/x² − 1, so dt = −2/x³ dx, i.e. dx/x³ = −dt/2.
4. When x = 1/3, t = 8; when x = 1, t = 0.
5. The integral becomes (1/2) ∫₀⁸ t^(1/3) dt = (1/2)(3/4) t^(4/3), evaluated from 0 to 8.
6. 8^(4/3) = 16, so the value is (3/8)(16) = 6, option (A).
7. The answer key at the back of the book prints (D) 4, but putting t = 1/x² – 1 turns the integral into ½ ∫₀⁸ ∛t dt = 6, so the answer is (A) 6.

Answer6, option (A)

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1. By the Fundamental Theorem of Calculus, if f(x) = ∫₀ˣ g(t) dt, then f′(x) = g(x).
2. Here g(t) = t sin t, so f′(x) is obtained by putting x in place of t.
3. f′(x) = x sin x, which is option (B).

Answerx sin x, option (B)

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