Exercise 7.2 answers: Integrals

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1. Look at the bottom, 1 + x². Its own rate of change is 2x, and that is exactly the top.
2. So put t = 1 + x². Then dt = 2x dx, which is the whole numerator.
3. The integral becomes ∫ dt/t = log|t| + C.
4. Since 1 + x² is always positive, drop the modulus: log(1 + x²) + C.

Answerlog(1 + x²) + C

Watch this explained “Substituting for the denominator”, 7:14 into Changing the variable until what is left is a standard form

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1. The rate of change of log x is 1/x, and 1/x is exactly the other factor here.
2. So put t = log x. Then dt = dx/x.
3. The integral becomes ∫ t² dt = t³/3 + C.
4. Put log x back for t: (log x)³/3 + C.

Answer(log x)³/3 + C

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1. Factor the bottom: x + x log x = x(1 + log x).
2. So the integral is ∫ 1/[x(1 + log x)] dx.
3. The rate of change of (1 + log x) is 1/x, which is exactly the remaining factor.
4. Put t = 1 + log x, so dt = dx/x, and the integral becomes ∫ dt/t = log|t| + C.
5. Put back t: log|1 + log x| + C.

Answerlog|1 + log x| + C

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1. The inner function here is cos x, and its rate of change is −sin x. The factor sin x standing outside is that rate with the sign changed.
2. Put t = cos x, so dt = −sin x dx, which means sin x dx = −dt.
3. The integral becomes −∫ sin t dt = −(−cos t) + C = cos t + C.
4. Put back t = cos x: cos(cos x) + C.

Answercos(cos x) + C

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1. The rate of change of sin(ax + b) is a cos(ax + b), which is (up to the constant a) the other factor here.
2. Put t = sin(ax + b), so dt = a cos(ax + b) dx.
3. The integral becomes (1/a) ∫ t dt = t²/(2a) + C.
4. Put back t: sin²(ax + b)/(2a) + C.

Answersin²(ax + b)/(2a) + C

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1. Put t = ax + b. Its rate of change is the constant a, so dt = a dx, which means dx = dt/a.
2. The integral becomes (1/a) ∫ t^1/2 dt = (1/a) · (2/3) t^3/2 + C.
3. Put back t = ax + b: 2(ax + b)^3/2/(3a) + C.

Answer2(ax + b)^3/2/(3a) + C

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1. Put t = x + 2, so x = t − 2 and dx = dt.
2. Rewrite the integrand in t: (t − 2)√t = t^3/2 − 2t^1/2.
3. Integrate: (2/5)t^5/2 − (4/3)t^3/2 + C.
4. Put back t = x + 2: (2/5)(x+2)^5/2 − (4/3)(x+2)^3/2 + C.

Answer(2/5)(x+2)^5/2 − (4/3)(x+2)^3/2 + C

Watch the lesson Changing the variable until what is left is a standard form

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1. The rate of change of 1 + 2x² is 4x, which is a constant multiple of the x sitting outside the root.
2. Put t = 1 + 2x², so dt = 4x dx, which means x dx = dt/4.
3. The integral becomes (1/4) ∫ √t dt = (1/4)(2/3)t^3/2 + C.
4. Put back t: (1/6)(1+2x²)^3/2 + C.

Answer(1/6)(1 + 2x²)^3/2 + C

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1. The rate of change of x² + x + 1 is 2x + 1, and 4x + 2 is exactly twice that.
2. Put t = x² + x + 1, so dt = (2x+1) dx, and (4x+2) dx = 2 dt.
3. The integral becomes 2∫ √t dt = 2 · (2/3) t^3/2 + C.
4. Put back t: (4/3)(x²+x+1)^3/2 + C.

Answer(4/3)(x² + x + 1)^3/2 + C

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1. Factor √x out of the bottom: x − √x = √x(√x − 1).
2. Put t = √x, so x = t² and dx = 2t dt.
3. The integral becomes ∫ 2t dt/[t(t−1)] = ∫ 2 dt/(t−1) = 2 log|t−1| + C.
4. Put back t = √x: 2 log|√x − 1| + C.

Answer2 log|√x − 1| + C

Watch the lesson Changing the variable until what is left is a standard form

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1. Put t = x + 4, so x = t − 4 and dx = dt.
2. Rewrite in t: (t−4)/√t = t^1/2 − 4t^-1/2.
3. Integrate: (2/3)t^3/2 − 8t^1/2 + C.
4. Put back t = x + 4: (2/3)(x+4)^3/2 − 8√(x+4) + C.

Answer(2/3)(x+4)^3/2 − 8√(x+4) + C

Watch the lesson Changing the variable until what is left is a standard form

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1. Split x⁵ = x³ · x². The rate of change of x³ − 1 is 3x², a constant multiple of the x² kept aside.
2. Put t = x³ − 1, so dt = 3x² dx, which means x² dx = dt/3. Also x³ = t + 1.
3. The integral becomes ∫ t^1/3(t + 1) · (dt/3) = (1/3) ∫ (t^4/3 + t^1/3) dt.
4. Integrate: (1/3)[(3/7)t^7/3 + (3/4)t^4/3] + C = (1/7)t^7/3 + (1/4)t^4/3 + C.
5. Put back t = x³ − 1: (1/7)(x³ − 1)^7/3 + (1/4)(x³ − 1)^4/3 + C.

Answer(1/7)(x³ − 1)^7/3 + (1/4)(x³ − 1)^4/3 + C

Watch the lesson Changing the variable until what is left is a standard form

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1. The rate of change of 2 + 3x³ is 9x², a constant multiple of the x² on top.
2. Put t = 2 + 3x³, so dt = 9x² dx, meaning x² dx = dt/9.
3. The integral becomes (1/9)∫ t⁻³ dt = (1/9)·t⁻²/(−2) + C = −1/(18t²) + C.
4. Put back t: −1/[18(2+3x³)²] + C.

Answer−1/[18(2 + 3x³)²] + C

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1. Integrate: 1/[x(log x)^m], x > 0, m ≠ 1.
2. The rate of change of log x is 1/x, which is exactly the other factor here.
3. Put t = log x, so dt = dx/x.
4. The integral becomes ∫ t^−m dt = t^1−m/(1−m) + C, valid since m ≠ 1.
5. Put back t: (log x)^1−m/(1−m) + C.

Answer(log x)^1−m/(1 − m) + C

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1. The rate of change of 9 − 4x² is −8x, a constant multiple of the x on top.
2. Put t = 9 − 4x², so dt = −8x dx, meaning x dx = −dt/8.
3. The integral becomes −(1/8)∫ dt/t = −(1/8) log|t| + C.
4. Put back t: −(1/8) log|9 − 4x²| + C.

Answer−(1/8) log|9 − 4x²| + C

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1. Put t = 2x + 3. Its rate of change is the constant 2, so dt = 2 dx.
2. The integral becomes (1/2)∫ e^t dt = (1/2)e^t + C.
3. Put back t: (1/2)e^2x+3 + C.

Answer(1/2)e^2x+3 + C

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1. Write the integrand as x·e^−x².
2. The rate of change of −x² is −2x, a constant multiple of the x standing there.
3. Put t = −x², so dt = −2x dx, meaning x dx = −dt/2.
4. The integral becomes −(1/2)∫ e^t dt = −(1/2)e^t + C.
5. Put back t: −(1/2)e^−x² + C, which is the same as −1/(2e^x²) + C.

Answer−1/(2e^x²) + C

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1. The rate of change of tan⁻¹x is 1/(1+x²), exactly the other factor here.
2. Put t = tan⁻¹x, so dt = dx/(1+x²).
3. The integral becomes ∫ e^t dt = e^t + C.
4. Put back t: e^tan−1x + C.

Answere^tan−1x + C

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1. Divide top and bottom by e^x: the integrand becomes (e^x − e^−x)/(e^x + e^−x).
2. Now notice the rate of change of the bottom, e^x + e^−x, is e^x − e^−x — exactly the top.
3. Put t = e^x + e^−x, so dt = (e^x − e^−x) dx.
4. The integral becomes ∫ dt/t = log|t| + C.
5. Put back t, and drop the modulus since e^x+e^−x is always positive: log(e^x + e^−x) + C.

Answerlog(e^x + e^−x) + C

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1. The rate of change of e^2x + e^−2x is 2e^2x − 2e^−2x, exactly twice the top.
2. Put t = e^2x + e^−2x, so dt = 2(e^2x − e^−2x) dx.
3. The integral becomes (1/2)∫ dt/t = (1/2) log|t| + C.
4. Put back t, and drop the modulus since e^2x+e^−2x is always positive: (1/2) log(e^2x + e^−2x) + C.

Answer(1/2) log(e^2x + e^−2x) + C

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1. tan²θ = sec²θ − 1, so write tan²(2x − 3) as sec²(2x − 3) − 1.
2. The integral splits into ∫sec²(2x − 3) dx − ∫1 dx.
3. ∫sec²(2x − 3) dx = (1/2) tan(2x − 3), because the 2 inside the bracket needs a matching 1/2 outside.
4. ∫1 dx = x.

Answer(1/2) tan(2x − 3) − x + C

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1. This is a standard form: ∫sec²(linear expression) dx = (1/slope) tan(linear expression).
2. Here the slope of 7 − 4x is −4, so divide by −4.

Answer−(1/4) tan(7 − 4x) + C

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1. Let t = sin^-1x. Then dt = dx/√(1 − x²), which is exactly the other factor in the integral.
2. The integral becomes ∫t dt = t²/2.
3. Put t back: (sin^-1x)²/2.

Answer(sin^-1x)²/2 + C

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1. The derivative of the denominator 6cos x + 4sin x is −6sin x + 4cos x.
2. Half of that derivative is −3sin x + 2cos x, which is exactly the numerator 2cos x − 3sin x.
3. So the numerator equals (1/2) times the denominator's own derivative.
4. An integral of the form (derivative of D)/D dx equals log|D|, so this integral is (1/2) log|denominator|.

Answer(1/2) log|6cos x + 4sin x| + C

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1. Let t = tan x. Then dt = sec²x dx = dx/cos²x, which is exactly the 1/cos²x factor.
2. The integral becomes ∫dt/(1 − t)².
3. ∫(1 − t)^-2 dt = 1/(1 − t).
4. Put t back: 1/(1 − tan x).

Answer1/(1 − tan x) + C

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1. Let u = √x. Then du = dx/(2√x), so dx/√x = 2 du.
2. The integral becomes ∫cos u · 2 du = 2 sin u.
3. Put u back: 2 sin(√x).

Answer2 sin(√x) + C

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1. Let u = sin 2x. Then du = 2cos 2x dx, so cos 2x dx = du/2.
2. The integral becomes ∫√u · (du/2) = (1/2)·(2/3) u^3/2 = (1/3) u^3/2.
3. Put u back: (1/3) (sin 2x)^3/2.

Answer(1/3) (sin 2x)^3/2 + C

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1. Let u = 1 + sin x. Then du = cos x dx, which is exactly the numerator.
2. The integral becomes ∫u^-1/2 du = 2u^1/2.
3. Put u back: 2√(1 + sin x).

Answer2√(1 + sin x) + C

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1. Let u = log(sin x). Then du = cot x dx, which is exactly the other factor.
2. The integral becomes ∫u du = u²/2.
3. Put u back: (log sin x)²/2.

Answer(log sin x)²/2 + C

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1. Let u = 1 + cos x. Then du = −sin x dx, so sin x dx = −du.
2. The integral becomes ∫(−du)/u = −log|u|.
3. Put u back: −log|1 + cos x|.

Answer−log|1 + cos x| + C

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1. Let u = 1 + cos x. Then du = −sin x dx, so sin x dx = −du.
2. The integral becomes ∫(−du)/u² = 1/u.
3. Put u back: 1/(1 + cos x).

Answer1/(1 + cos x) + C

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1. Write 1/(1 + cot x) as sin x/(sin x + cos x).
2. The derivative of sin x + cos x is cos x − sin x.
3. Split sin x as half of (sin x + cos x) minus half of (cos x − sin x).
4. This splits the integral into (1/2)∫dx minus (1/2)∫(derivative of denominator)/denominator dx.
5. That gives x/2 minus (1/2) log|sin x + cos x|.

Answerx/2 − (1/2) log|sin x + cos x| + C

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1. Write 1/(1 − tan x) as cos x/(cos x − sin x).
2. The derivative of cos x − sin x is −sin x − cos x.
3. Split cos x as half of (cos x − sin x) plus half of (sin x + cos x).
4. This splits the integral into (1/2)∫dx plus (1/2)∫(sin x + cos x)/(cos x − sin x) dx, and that second part works out to −(1/2) log|cos x − sin x|.

Answerx/2 − (1/2) log|cos x − sin x| + C

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1. Let t = tan x, so sin x cos x = t/(1 + t²) and dx = dt/(1 + t²).
2. Substituting, the (1 + t²) cancels and the integral becomes ∫√t/t dt = ∫t^-1/2 dt.
3. ∫t^-1/2 dt = 2t^1/2.
4. Put t back: 2√(tan x).

Answer2√(tan x) + C

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1. Let u = 1 + log x. Then du = dx/x, which is exactly the other factor.
2. The integral becomes ∫u² du = u³/3.
3. Put u back: (1 + log x)³/3.

Answer(1 + log x)³/3 + C

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1. Write (x + 1)/x as 1 + 1/x.
2. Let u = x + log x. Then du = (1 + 1/x) dx, which is exactly that factor.
3. The integral becomes ∫u² du = u³/3.
4. Put u back: (x + log x)³/3.

Answer(x + log x)³/3 + C

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1. Let t = x⁴. Then dt = 4x³ dx, so x³ dx = dt/4. Also 1 + x⁸ = 1 + t².
2. The integral becomes (1/4)∫sin(tan^-1t)/(1 + t²) dt.
3. Let φ = tan^-1t. Then dφ = dt/(1 + t²), which is exactly the other factor.
4. The integral becomes (1/4)∫sin φ dφ = −(1/4) cos φ.
5. Since cos(tan^-1t) = 1/√(1 + t²), this simplifies to −1/(4√(1 + t²)).
6. Put t back: −1/(4√(1 + x⁸)).

Answer−1/(4√(1 + x⁸)) + C

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1. Question: ∫(10x⁹ + 10ˣ logₑ10 dx)/(x¹⁰ + 10ˣ).
2. The derivative of x¹⁰ + 10^x is 10x⁹ + 10^x log_e10, which is exactly the numerator.
3. So the integral has the form (derivative of D)/D dx, which integrates to log|D|.
4. That matches option (D).

Answer(D) log(10ˣ + x¹⁰) + C

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1. 1/(sin²x cos²x) = (sin²x + cos²x)/(sin²x cos²x), using sin²x + cos²x = 1.
2. Split the fraction: sin²x/(sin²x cos²x) + cos²x/(sin²x cos²x) = 1/cos²x + 1/sin²x = sec²x + cosec²x.
3. ∫sec²x dx = tan x and ∫cosec²x dx = −cot x.
4. Adding the two: tan x − cot x, which matches option (B).

Answer(B) tan x − cot x + C

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