Exercise 7.6 answers: Integrals

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1. This is a product, so use integration by parts.
2. x gets smaller when differentiated (it becomes 1), so let x be the first function and sin x the second.
3. By parts: ∫x sin x dx = x·(−cos x) − ∫1·(−cos x) dx.
4. = −x cos x + ∫cos x dx = −x cos x + sin x + C.

Answer−x cos x + sin x + C

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1. x is a power, so it should be the first function (it shrinks under differentiation); sin 3x is the second.
2. ∫sin 3x dx = −cos 3x/3.
3. By parts: x·(−cos3x/3) − ∫1·(−cos3x/3) dx = −x cos3x/3 + (1/3)∫cos3x dx.
4. = −x cos3x/3 + (1/3)(sin3x/3) + C = −x cos3x/3 + sin3x/9 + C.

Answer−x cos 3x/3 + sin 3x/9 + C

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1. x² is a power (it shrinks to x, then to 1, then to 0 under repeated differentiation), so make it the first function; eˣ is the second, since eˣ integrates to itself.
2. First pass: ∫x²eˣ dx = x²eˣ − ∫2x eˣ dx.
3. Second pass: ∫2x eˣ dx = 2x eˣ − ∫2 eˣ dx = 2x eˣ − 2eˣ.
4. Combine: x²eˣ − (2x eˣ − 2eˣ) + C = eˣ(x² − 2x + 2) + C.

Answereˣ(x² − 2x + 2) + C

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1. log x cannot be integrated directly, so it is the first function; x is the second.
2. ∫x dx = x²/2.
3. By parts: (log x)(x²/2) − ∫(1/x)(x²/2) dx = (x²/2) log x − (1/2)∫x dx.
4. = (x²/2) log x − (1/2)(x²/2) + C = (x²/2) log x − x²/4 + C.

Answer(x²/2) log x − x²/4 + C

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1. log 2x cannot be integrated directly, so it is the first function; x is the second.
2. Differentiating log 2x gives 1/x (the 2 cancels), and ∫x dx = x²/2.
3. By parts: (x²/2) log 2x − ∫(1/x)(x²/2) dx = (x²/2) log 2x − (1/2)∫x dx.
4. = (x²/2) log 2x − x²/4 + C.

Answer(x²/2) log 2x − x²/4 + C

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1. log x cannot be integrated directly, so it is the first function; x² is the second.
2. ∫x² dx = x³/3.
3. By parts: (log x)(x³/3) − ∫(1/x)(x³/3) dx = (x³/3) log x − (1/3)∫x² dx.
4. = (x³/3) log x − (1/3)(x³/3) + C = (x³/3) log x − x³/9 + C.

Answer(x³/3) log x − x³/9 + C

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1. sin⁻¹x cannot be integrated directly, so it is the first function; x is the second, and ∫x dx = x²/2.
2. By parts: ∫x sin⁻¹x dx = (x²/2) sin⁻¹x − ∫[1/√(1−x²)]·(x²/2) dx = (x²/2) sin⁻¹x − (1/2)∫x²/√(1−x²) dx.
3. Write x² = 1 − (1−x²), so x²/√(1−x²) = 1/√(1−x²) − √(1−x²).
4. ∫1/√(1−x²) dx = sin⁻¹x, and ∫√(1−x²) dx = (x/2)√(1−x²) + (1/2) sin⁻¹x.
5. So ∫x²/√(1−x²) dx = sin⁻¹x − (x/2)√(1−x²) − (1/2) sin⁻¹x = (1/2) sin⁻¹x − (x/2)√(1−x²).
6. Put this back: (x²/2) sin⁻¹x − (1/2)[(1/2) sin⁻¹x − (x/2)√(1−x²)] + C = (x²/2) sin⁻¹x − (1/4) sin⁻¹x + (x/4)√(1−x²) + C.
7. Collect the sin⁻¹x terms: [(2x²−1)/4] sin⁻¹x + (x√(1−x²))/4 + C.

Answer[(2x²−1)/4] sin⁻¹x + (x√(1−x²))/4 + C

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1. tan⁻¹x cannot be integrated directly, so it is the first function; x is the second.
2. ∫x dx = x²/2. By parts: (tan⁻¹x)(x²/2) − ∫[1/(1+x²)]·(x²/2) dx.
3. x²/(1+x²) = 1 − 1/(1+x²), so ∫x²/(1+x²) dx = x − tan⁻¹x.
4. So the integral is (x²/2) tan⁻¹x − (1/2)(x − tan⁻¹x) + C = [(x²+1)/2] tan⁻¹x − x/2 + C.

Answer[(x²+1)/2] tan⁻¹x − x/2 + C

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1. cos⁻¹x cannot be integrated directly, so it is the first function; x is the second, and ∫x dx = x²/2. The rate of change of cos⁻¹x is −1/√(1−x²).
2. By parts: ∫x cos⁻¹x dx = (x²/2) cos⁻¹x − ∫[−1/√(1−x²)]·(x²/2) dx = (x²/2) cos⁻¹x + (1/2)∫x²/√(1−x²) dx.
3. Write x² = 1 − (1−x²), so ∫x²/√(1−x²) dx = ∫1/√(1−x²) dx − ∫√(1−x²) dx = sin⁻¹x − [(x/2)√(1−x²) + (1/2) sin⁻¹x] = (1/2) sin⁻¹x − (x/2)√(1−x²).
4. So the integral is (x²/2) cos⁻¹x + (1/4) sin⁻¹x − (x/4)√(1−x²) + C₁.
5. Use sin⁻¹x = π/2 − cos⁻¹x: (1/4) sin⁻¹x = π/8 − (1/4) cos⁻¹x, and the number π/8 joins the constant.
6. Collect the cos⁻¹x terms: [(2x²−1)/4] cos⁻¹x − (x√(1−x²))/4 + C.

Answer[(2x²−1)/4] cos⁻¹x − (x√(1−x²))/4 + C

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1. There is only one function here, so write it as (sin⁻¹x)² times 1; let (sin⁻¹x)² be the first function and 1 the second.
2. ∫1 dx = x. By parts: x(sin⁻¹x)² − ∫x · 2 sin⁻¹x · [1/√(1−x²)] dx.
3. For ∫2x sin⁻¹x/√(1−x²) dx, integrate by parts again: let sin⁻¹x be first, and x/√(1−x²) second, which integrates to −√(1−x²).
4. That gives 2[−√(1−x²) sin⁻¹x + ∫1 dx] = −2√(1−x²) sin⁻¹x + 2x.
5. So the whole integral is x(sin⁻¹x)² − (−2√(1−x²) sin⁻¹x + 2x) + C = x(sin⁻¹x)² + 2√(1−x²) sin⁻¹x − 2x + C.

Answerx(sin⁻¹x)² + 2√(1−x²) sin⁻¹x − 2x + C

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1. cos⁻¹x cannot be integrated on its own, so it is the first function; the rest, x/√(1−x²), is the second.
2. That second factor integrates by substitution: put u = 1−x², du = −2x dx, giving ∫x/√(1−x²) dx = −√(1−x²).
3. By parts: (cos⁻¹x)(−√(1−x²)) − ∫[−1/√(1−x²)]·(−√(1−x²)) dx = −√(1−x²) cos⁻¹x − ∫1 dx.
4. = −√(1−x²) cos⁻¹x − x + C.
5. The answer key at the back of the book prints –√(1–x²) cos⁻¹x + x + C, but differentiating that gives the question's function plus 2, so the sign of x is a slip and the answer is –√(1–x²) cos⁻¹x – x + C.

Answer−√(1 − x²) cos⁻¹x − x + C

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1. x is a power, so it is the first function; sec²x is the second, since sec²x integrates to tan x.
2. By parts: x tan x − ∫1·tan x dx = x tan x − ∫tan x dx.
3. ∫tan x dx = −log|cos x|, so subtracting it back gives + log|cos x|.
4. So the integral is x tan x + log|cos x| + C.

Answerx tan x + log|cos x| + C

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1. There is only one function here, so write tan⁻¹x as tan⁻¹x times 1; let tan⁻¹x be the first function and 1 the second.
2. ∫1 dx = x. By parts: x tan⁻¹x − ∫x·[1/(1+x²)] dx.
3. For ∫x/(1+x²) dx, put t = 1+x², dt = 2x dx, giving (1/2) log(1+x²).
4. So the integral is x tan⁻¹x − (1/2) log(1+x²) + C.

Answerx tan⁻¹x − (1/2) log(1+x²) + C

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1. (log x)² cannot be integrated directly, so it is the first function; x is the second.
2. ∫x dx = x²/2. By parts: (log x)²·(x²/2) − ∫[2 log x/x]·(x²/2) dx = (x²/2)(log x)² − ∫x log x dx.
3. ∫x log x dx was found in question 4: (x²/2) log x − x²/4.
4. So the integral is (x²/2)(log x)² − [(x²/2) log x − x²/4] + C = (x²/2)(log x)² − (x²/2) log x + x²/4 + C.

Answer(x²/2)(log x)² − (x²/2) log x + x²/4 + C

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1. This is a product of x² + 1 and log x, so use integration by parts.
2. log x cannot be integrated directly but is easy to differentiate — so it takes the differentiating role, and x² + 1 takes the integrating role.
3. ∫(x² + 1)dx = x³/3 + x, so by parts: ∫(x²+1)log x dx = (x³/3 + x) log x − ∫(x³/3 + x) · (1/x) dx.
4. Simplify inside the remaining integral: (x³/3 + x)/x = x²/3 + 1.
5. Integrate that: ∫(x²/3 + 1)dx = x³/9 + x.
6. So the integral is (x³/3 + x) log x − x³/9 − x + C.

Answer(x³/3 + x) log x − x³/9 − x + C

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1. Notice the bracket holds sin x and its own derivative cos x — this is the eˣ[f(x) + f′(x)] shape.
2. Here f(x) = sin x, and f′(x) = cos x, which matches exactly.
3. For this shape, the integral is simply eˣ f(x).
4. So ∫eˣ(sin x + cos x)dx = eˣ sin x + C.

Answereˣ sin x + C

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1. Rewrite the fraction: x/(1+x)² = (1 + x − 1)/(1+x)² = 1/(1+x) − 1/(1+x)².
2. So the integrand is eˣ[1/(1+x) − 1/(1+x)²].
3. Let f(x) = 1/(1+x). Then f′(x) = −1/(1+x)², so the bracket is exactly f(x) + f′(x).
4. This is the eˣ[f(x)+f′(x)] shape, so the integral is eˣ f(x).
5. So ∫x eˣ/(1+x)² dx = eˣ/(1+x) + C.

Answereˣ⁄(1 + x) + C

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1. Use half-angle formulas: 1 + cos x = 2cos²(x/2), and sin x = 2 sin(x/2)cos(x/2).
2. So (1+sin x)/(1+cos x) = 1/(2cos²(x/2)) + sin(x/2)/cos(x/2) = (1/2)sec²(x/2) + tan(x/2).
3. Let f(x) = tan(x/2). Then f′(x) = (1/2)sec²(x/2), so the bracket is f(x) + f′(x).
4. This is the eˣ[f(x)+f′(x)] shape, so the integral is eˣ f(x).
5. So the integral is eˣ tan(x/2) + C.

Answereˣ tan(x⁄2) + C

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1. Let f(x) = 1/x. Then f′(x) = −1/x², so the bracket 1/x − 1/x² is f(x) + f′(x).
2. This is the eˣ[f(x)+f′(x)] shape, so the integral is eˣ f(x).
3. So ∫eˣ(1/x − 1/x²)dx = eˣ/x + C.

Answereˣ⁄x + C

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1. Rewrite: (x−3)/(x−1)³ = (x−1−2)/(x−1)³ = 1/(x−1)² − 2/(x−1)³.
2. Let f(x) = 1/(x−1)². Then f′(x) = −2/(x−1)³, so the bracket is f(x) + f′(x).
3. This is the eˣ[f(x)+f′(x)] shape, so the integral is eˣ f(x).
4. So the integral is eˣ⁄(x−1)² + C.

Answereˣ⁄(x − 1)² + C

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1. Neither factor becomes simpler on differentiating, so apply integration by parts and keep going.
2. Let u = sin x, dv = e^2x dx, so du = cos x dx, v = e^2x/2.
3. ∫e^2x sin x dx = (e^2x sin x)/2 − (1/2)∫e^2x cos x dx.
4. Apply parts again to ∫e^2x cos x dx with u = cos x, dv = e^2xdx: this gives (e^2x cos x)/2 + (1/2)∫e^2x sin x dx.
5. Substitute back: I = (e^2x sin x)/2 − (e^2x cos x)/4 − (1/4)I, where I is the original integral.
6. Collect I: (5/4)I = e^2x(2 sin x − cos x)/4, so I = e^2x(2 sin x − cos x)/5.
7. So the integral is e^2x(2 sin x − cos x)/5 + C.

Answere^2x(2 sin x − cos x)⁄5 + C

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1. For x between −1 and 1, sin⁻¹(2x/(1+x²)) equals 2 tan⁻¹x (a standard identity), so integrate 2tan⁻¹x instead.
2. Integrate by parts: let u = tan⁻¹x, dv = dx, so du = 1/(1+x²) dx, v = x.
3. ∫tan⁻¹x dx = x tan⁻¹x − ∫x/(1+x²) dx = x tan⁻¹x − (1/2)log(1+x²).
4. Multiply by 2: ∫2tan⁻¹x dx = 2x tan⁻¹x − log(1+x²).
5. So the integral is 2x tan⁻¹x − log(1+x²) + C.

Answer2x tan⁻¹x − log(1 + x²) + C

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1. Question: ∫x²e^(x³)dx.
2. The derivative of x³ is 3x², and x² is already in the integrand — a signal to substitute t = x³.
3. Put t = x³, so dt = 3x² dx, which means x² dx = dt⁄3.
4. The integral becomes ∫e^t · (dt⁄3) = (1⁄3)e^t + C.
5. Put t back as x³: (1⁄3)e^x³ + C.
6. Check by differentiating: (1⁄3)e^x³ · 3x² = x²e^x³, the integrand. This is option (A).

Answer(A) (1⁄3)e^x³ + C

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1. Expand the bracket: eˣ sec x(1+tan x) = eˣ[sec x + sec x tan x].
2. Let f(x) = sec x. Then f′(x) = sec x tan x, so the bracket is f(x) + f′(x).
3. This is the eˣ[f(x)+f′(x)] shape, so the integral is eˣ f(x) = eˣ sec x + C.
4. This matches option (B).

Answer(B) eˣ sec x + C

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